Important note. All references to ‘For At Least One Thing:’ FALOT’ have been replaced with ‘There is something such that:’ ‘TISS’ Important Change.

Important note. All references to ‘For At Least One Thing:’ FALOT’ have been replaced with ‘There is something such that:’ ‘TISS’ Important Change

For each and every thing in the universe: For all things:
For each thing: For each and every thing in the universe: For all things: For anything: For everything in the universe (call it x): For every x: For anything you pick: These are some of the ways to express a universal statement. When a statement starts like one of these it takes many specific objects to prove it true but only one object to prove it false. We will use the “For Each Thing:” phrase as our standard form (and abbreviate it “FET”), but if another phrase sounds better in expressing a particular statement, feel free to think of the statement in those terms. You can put “in the universe” on the end of anything that doesn’t already have it, if you want. Think about this and then go forward with the right arrow. Explanation

There is something such that:
For at least one thing: For something: There is something such that: There is at least one thing in the universe such that: There is at least one thing (call it x): For some x: I can find something such that: These are some of the ways to express an existential statement. When a statement starts like one of these it takes many specific objects to prove it false but only one object to prove it true. We will use the “There is something such that:” phrase as our standard form (which we will abbreviate “TISS”) , but if another sounds better in expressing a particular statement, feel free to think in those terms. You may put “in the universe such that” on the end of anything above that doesn’t already have it, if you want. Think about this and then go forward for some other hints. Explanation

Ron McIntyre's Basic Forms
Basic Form for 'All A's are B's': (x)(Ax  Bx). Some examples: All cats are friendly. (Equivalents: Every cat is friendly; any cat is friendly; cats are all friendly; cats are always friendly.) For every x, if x is a cat then x is friendly. For every x, if Cx then Fx. (x)(Cx  Fx) . All domesticated cats are friendly. For every x, if x is domesticated and x is a cat, then x is friendly. For every x, if Dx and Cx, then Fx. (x)[(Dx & Cx)  Fx] . All cats that are treated well are friendly. For every x, if x is a cat and x is treated well, then x is friendly. (x)[(Cx & Tx)  Fx] . All cats are either friendly or aloof. For every x, if x is a cat then either x is friendly or x is aloof. (x)[Cx  (Fx v Ax)] . Basic Form for 'No A's are B's': (x)(Ax  ~Bx) or {Equivalent: ~(x)(Ax & Bx) } No tigers are friendly. {Equivalents: All tigers are unfriendly; tigers are never friendly.) For every x, if x is a tiger then x is not friendly. For every x, if Tx then ~Fx. (x)(Tx  ~Fx) . {Equivalent: ~(x)(Tx & Fx) } Forward for more Ron McIntyre's Basic Forms

No wild tigers are friendly. For every x, if x is wild and x is a tiger, then x is not friendly. (x)[(Wx & Tx)  ~Fx] . {Equivalent: ~(x)((Wx & Tx) & Fx) .} No tigers are contented pets. For every x, if x is a tiger then it is not the case that x is contented and x is a pet. (x)[Tx  ~(Cx & Px)] .{Equivalent: ~(x)(Tx & (Cx & Px)) } No tigers are safe if they are uncaged. For every x, if x is a tiger, then if x is not caged x is not safe. (x)[Tx  (~Cx  ~Sx)] {Equivalent: ~(x)(Tx & (Sx & ~Cx)) } Basic Form for 'Some A's are B's': (x)(Ax & Bx) . Some examples: Some cats are friendly. {Equivalents: There are friendly cats; cats are sometimes friendly; not all cats are unfriendly.) For at least one x, x is a cat and x is friendly. For at least one x, Cx and Fx. (x)(Cx & Fx) . Some domesticated cats are friendly. For at least one x, x is domesticated and x is a cat and x is friendly. (x)[(Dx & Cx) & Fx] . Some cats are not friendly. For at least one x, x is a cat and x is not friendly. (x)(Cx & ~Fx) . Forward for more. Ron McIntyre's Basic Forms

Some cats are neither friendly nor pretty. For at least one x, x is a cat and x is neither friendly nor pretty. For at least one x, x is a cat and x is not friendly and x is not pretty. (x)[Cx & (~Fx & ~Px)] . Basic Form for 'Only A's are B's': (x)(Bx  Ax) {Equivalent form: (x)(~Ax  ~Bx) } Some examples: Only women are mothers. (Equivalent: None but women are mothers.} For every x, x is a mother only if x is a woman. For every x, if x is a mother then x is a woman [or: if x is not a woman then x is not a mother]. (x)(Mx  Wx) . {Equivalent: (x)(~Wx  ~Mx) .} Only adult women are mothers. For every x, x is a mother only if x is an adult and x is a woman. (x)[Mx  (Ax & Wx)] . {Equivalent: (x)[~(Ax & Wx)  ~Mx]. or (x)[(~Ax v ~Wx)  ~Mx] .} Only women are mothers or sisters. For every x, x is a mother or x is a sister only if x is a woman. (x)[(Mx v Sx)  Wx] . {Equivalent: (x)[~Wx  ~(Mx v Sx)]. or (x)[~Wx  (~Mx & ~Sx)] .} NOTE: 'The only A's are B's' is not equivalent to 'Only A's are B's' but to 'All A's are B's'. Thus, the correct translation of 'The only people who are mothers are women' is '(x)[(Px & Mx)  Wx]'. NOTE: The basic form for 'A's and B's are C's' is not '(x)[(Ax & Bx)  Cx]', but '(x)[(Ax v Bx)  Cx]'. Thus, a correct translation of 'Dogs and cats are animals' is '(x)[(Dx v Cx)  Ax]'. But it could also be handled as an ampersand statement. i.e.; All Dogs are animals AND all cats are animals: ((x)(Dx  Ax) & (x)(Cx  Ax)) . Go forward for first example. Ron McIntyre's Basic Forms

Doctors are Professionals
First we should try to figure out what this means. Does it mean All Doctors are professionals or some Doctors? It seems to be “All”. So that gives us a hint as to what standard form to use Think about that and go forward with the right arrow. Problem 1

Doctors are Professionals For Each Thing:….
It is a “For Each Thing” kind of statement. But what should happen now? It cannot be “For Each Thing: it is a doctor and it is a professional” for that would mean that everything in the universe has the property of being both a doctor and a professional, including you. But ask yourself: if this statement were true and you found a Doctor, then what would you be able to conclude about him? If he were a doctor then he would have to be a professional. So imagine completing the red line and then hit the right arrow. Problem 1

For Each Thing: If it is a Doctor then it is a Professional. Now that our target statement is in the ‘For Each Thing:” standard form, we use an individual’s name to represent any old thing. Let us choose “Alan” and let us delete the major connector “For Each Thing”. Hit the forward arrow to continue. Problem 1

For Each Thing: If it is a Doctor then it is a Professional. If Alan is a Doctor then Alan is a Professional. Now we see that the new statement is also in standard form, so we can break it into its parts. Imagine and then go forward. Problem 1

For Each Thing: If it is a Doctor then it is a Professional. If Alan is a Doctor then Alan is a Professional. Alan is a Doctor Alan is a Professional. “Alan is a Doctor” is not in standard form and has no standard form synonym and neither does “Alan is a Professional” and so let us use “D_” for “_ is a doctor” and “P_” for “_ is a professional” If that were the dictionary given, it would be an extra clue that we were now to represent atomics. Hit right arrow to go forward. Problem 1

For Each Thing: If it is a Doctor then it is a Professional. If Alan is a Doctor then Alan is a Professional. Alan is a Doctor Alan is a Professional. Da Pa So if “Da” represents one side of an “if..then” standard form and “Pa” the other then how will We represent the statement in green? Go forward. Problem 1

For Each Thing: If it is a Doctor then it is a Professional. If Alan is a Doctor then Alan is a Professional. (Da  Pa) Alan is a Doctor Alan is a Professional. Da Pa Right, all “If… Then” standard forms get represented with a horseshoe. But the statement in red came from the “For Each Thing” standard form, so how will we represent the statement In green? Hit forward Problem 1

For Each Thing: If it is a Doctor then it is a Professional. (x)(Dx  Px) If Alan is a Doctor then Alan is a Professional. (Da É Pa) Alan is a Doctor Alan is a Professional. Da Pa All “For Each Thing” statements are represented by universal quantifiers. “Alan” was our representative individual. So now “a” will be replaced by “x”. Reading it back into English we have: For Each thing (call it ‘x’) if x (it) has the property of being a Doctor then x has the property of being a professional. Variables take the role of “it” in English. The English statement in red is synonymous with the statement in green, so how will we represent that? Go forward Problem 1

(x)(Dx  Px) For Each Thing: If it is a Doctor then it is a Professional. If Alan is a Doctor then Alan is a Professional. (Da  Pa) Alan is a Doctor Alan is a Professional. Da Pa Synonymous statements are represented the same way. So there you have it. In the future we shall Shorthand “For Each Thing” statements as “FET”. Let us make that substitution now. Hit forward. Problem 1

(x)(Dx  Px) FET: If it is a Doctor then it is a Professional. If Alan is a Doctor then Alan is a Professional. (Da  Pa) Alan is a Doctor Alan is a Professional. Da Pa So “(•x)(Dx É Px)” is the answer. Hit forward for new problem. Problem 1

Doctors are in the Audience.
New Problem First we should try to figure out what this means. Does it mean “All Doctors are in the audience” or just Some Doctors? It seems to be “Some”. So that gives us a hint as to what standard form to use Think about that and go forward with the right arrow. Problem 2

Doctors are in the Audience. There is something such that:….
It is a “There is something such that:” kind of statement. But what should happen now? Think about what it would take to prove that “Doctors are in the audience.”? What kind of thing would I have to bring into your life to show it for certain? I’d have to find something, it would have to be a doctor and it would have to be in the audience. (Let us abbreviate “There is something such that” as “TISS”. That will be sort of a TISS symbol.) So imagine completing the red line and then hit the right arrow. Problem 2

TISS: it is a Doctor and it is in the Audience. Now that our target statement is in the ‘TISS:” standard form, we use an individual’s name to represent any old thing. Let us choose “Alan” again and let us delete the major connector. “TISS”. Hit the forward arrow to continue. Problem 2

TISS: it is a Doctor and it is in the Audience. Alan is a Doctor and Alan is in the Audience. So now we see that the new statement is also in standard form, so we can break it into its parts. Imagine and hit forward. Problem 2

TISS: it is a Doctor and it is in the Audience. Alan is a Doctor and Alan is in the Audience. Alan is a Doctor Alan is in the Audience. “Alan is a Doctor” is not in standard form and has no standard form synonym and neither does “Alan is in the Audience” and so let us use “D_” for “_ is a doctor” and “A_” for “_is in the Audience” If that were the dictionary given, it would be an extra clue that we were now to represent atomics. Imagine and hit forward Problem 2

TISS: it is a Doctor and it is in the Audience. Alan is a Doctor and Alan is in the Audience. Alan is a Doctor Alan is in the Audience. Da Aa So if “Da” represents one side of an “And” standard form and “Aa” the other then how will we represent the statement in green? Go forward. Problem 2

TISS: it is a Doctor and it is in the Audience. Alan is a Doctor and Alan is in the Audience. (Da & Aa) Alan is a Doctor Alan is in the Audience. Da Aa Right, all “And” standard forms get represented with an ampersand. But the statement in red came from the “TISS” standard form. So how will we represent the statement in green? Go forward. Problem 2

TISS: it is a Doctor and it is in the Audience. (x)(Da & Aa) Alan is a Doctor and Alan is in the Audience. (Da & Aa) Alan is a Doctor Alan is in the Audience. Da Aa All “TISS” statements are represented by existential quantifiers. “Alan” was our representative individual. So now “a” will be replaced by “x”. Reading it back into English we have: There is something: (call it ‘x’) x (it) has the property of being a Doctor and x has the property of being in the audience. Variables take the role of “it” in English. “There is something such that” could also be read as “There is at least one thing” or “I can find something such that”. The English statement in red is synonymous with the statement in green, so how will we represent that? Imagine and go forward. Problem 2

Doctors are in the Audience. (x)(Da & Aa)
TISS: it is a Doctor and it is in the Audience. Alan is a Doctor and Alan is in the Audience. (Da & Aa) Alan is a Doctor Alan is in the Audience. Da Aa Synonymous statements are represented the same way. So there you have it. The answer to turn in would be: “($x)(Da & Aa)” Hit forward for new problem. Problem 2

Only Doctors are in the Audience.
New Problem Consider the target statement in red. Is it in standard form? No. Can we restate it in standard form? Sure, but which form. If it could be shown true by one thing, then we would say that it is in the TISS form. But if it takes many things to prove it true or if one thing would prove it false, then it is a FET kind of statement. We can see that finding something in the audience that is not a Doctor would show it false. So it is a FET kind of statement. We know that FET statements usually have an “if then” form as the next connector. So how shall we proceed? Should we say “FET: if it is a Doctor then it is in the Audience.”, or should we say “FET: If it is in the Audience then it is a doctor.”? Hit forward to continue. Problem 3

FET: if it is in the Audience then it is a Doctor It cannot be “FET: If it is a Doctor then it is in the Audience.”, because that would mean “All Doctors in the world are in the Audience.” which is improbable, but we see that “Only Doctors are in the Audience” is not the kind of statement that is improbable. So let us use A_, and D_ as our dictionary and jump to how this statement in complete standard form should be represented. Hit Forward. Problem 3

FET: if it is in the Audience then it is a Doctor (x)(Dx  Ax) There you have the standard structure of a FET statement. A universal statement whose next major connector is a horseshoe. Since the statement in green is synonymous with the English statement in red then they will be represented the same way. Hit forward for new example. Problem 3

A doctor is devoted to the relief of suffering.
New Problem With this one again we should ask if it means to talk of all doctors or some doctors. From the ‘A’ it might seem clear that it is one doctor. But notice that this is a common way of referring to “all” things. Note that “A whale is a mammal” is not just talking about one whale, it is more to suggest that whales by their very nature are mammals. So we should take this to mean “All doctors are devoted to the relief of suffering.” So how would we say that as a “FET” statement? Think and then hit forward. Problem 4

FET:if it is a doctor then it is devoted to the relief of suffering This statement may seem odd since “devoted to the relief of suffering” may seem like a rather long property. But it’s ok for properties to be long. “Devoted to the relief of suffering” is something that an object can be. This form can now be turned into symbols with our level of experience. Think and then hit forward. Problem 4

FET:if it is a doctor then it is devoted to the relief of suffering (x)(Dx  Rx) An since this is synonymous with the green statement, it will be represented the same way. Think and then hit forward. Problem 4

A doctor is devoted to the relief of suffering. (x)(Dx  Rx)
FET:if it is a doctor then it is devoted to the relief of suffering So again we the classic form of a universal quantifier going with a horseshoe Hit forward for new problem. Problem 4

A doctor is devoted to my wife. New Problem
We won’t make the mistake of thinking that this is about all doctors, for it’s clearly not the case that doctors are by their very nature devoted to my wife. In a way we are using what is called ‘the principle of charity’. If there is an ambiguity and the statement could be true interpreted one way but is unlike to be true the other way of interpretation then choose the interpretation that could be true. Very charitable. So this says “there is at least one thing. It is a doctor and it is devoted to my wife”, for that indicates the kind of object it would take to prove the statement true. Think and then hit forward. Problem 5

A doctor is devoted to my wife.
TISS:it is a doctor and it is devoted to my wife. The us assume we can turn this into symbols in one step. Think and then hit forward. Problem 5

A doctor is devoted to my wife.
TISS:it is a doctor and it is devoted to my wife. (x)(Dx & Wx) And since it is synonymous with the green statement, the green will be the same. Think and then hit forward. Problem 5

A doctor is devoted to my wife. (x)(Dx & Wx)
TISS:it is a doctor and it is devoted to my wife. And since it is synonymous with the green statement, the green will be the same. Think and then hit forward. Problem 5

Doctors are not all rich.
New Problem Consider the target statement in red. What does it mean. In particular how many things would it take to prove it true or false? It looks like I could prove it true by finding one thing that was both a Doctor and not rich. So that tells me that it should be a TISS kind of statement. Hit forward. Problem 6

TISS: How should we finish this statement? Because we realize that we can prove this statement true with one thing, and we know what kind of thing it is. Spell out to yourself what kind of thing it is. Hit forward Problem 6

Doctors are not all rich. TISS: it is a Doctor and it is not rich.
It is a thing that is a Doctor and not rich. To say it is not rich is to say that it is not the case that it has the property of being rich. So we should now try to turn this into symbols since it is a familiar form. Hit forward. Problem 6

Doctors are not all rich. TISS: it is a Doctor and it is not rich.
(x)(Dx&~Rx) So this says “I can find something such that it is a doctor and it is not rich. But where did the “all” go, from the original statement? Well it turns out that there is another natural standard form synonym of “Doctors are not all rich”. T hink about what someone is denying when they say the statement in green. Hit forward. Problem 6

The statement in green denies that all doctors are rich. It is as if someone said “all doctors are rich” and the speaker said. “No, that’s not true.” So let us see the standard form version of the green as a “It is not the case that:” statement. Hit forward. Problem 6

Doctors are not all rich. NOT: All doctors are rich.
But since it is now in standard form, we can consider the statement it denies. Hit forward Problem 6

But the statement in red is in a form we have seen many times now. We can turn it into symbols immediately. (I hope). Hit forward. Problem 6

(x)(Dx  Rx) So, since this represents the English statement in red, how will be represent the “NOT” statement in green? Hit forward Problem 6

Doctors are not all rich. NOT: All doctors are rich. ~(x)(Dx  Rx)
Exactly right. And since the green statement is synonymous with the red English statement, then it can be represented the same way. So now we see that there are two ways to represent “Doctors are not all rich”. Hit forward. Problem 6

Doctors are not all rich. NOT: All doctors are rich. ~(x)(Dx  Rx)
TISS: it is a Doctor and it is not rich. (x)(Dx & ~Rx) How can these both be correct? Because they are logically equivalent. We will be able to show this when we get to quantified trees and derivations. Hit forward for new example. Problem 6

Imagine and hit forward.
No doctors are rich. New Problem Consider the statement in green. It is not in a standard form but could be restated in standard for. In fact there are two natural ways to approach this. First let us assess what it would take to show it true or false. If it takes one thing to prove it true then it would be an existential. If it takes one thing to prove it false then it would be a universal. It looks like one thing would prove it false. So let us try that. Imagine and hit forward. Problem 7

No doctors are rich. FET:…
If take this statement as true and we found a doctor could we conclude anything about his richness? It looks like we could. We could conclude that he was not rich. So that’s the way we should proceed. If he’s a doctor then he’s not rich. Hit forward. Problem 7

FET:if it is a doctor then it is not rich.
No doctors are rich. FET:if it is a doctor then it is not rich. This we should be able to turn into symbols because it is much like other “FET”statements. Think about it and hit forward. Problem 7

FET:if it is a doctor then it is not rich. (x)(Dx  ~Rx)
No doctors are rich. FET:if it is a doctor then it is not rich. (x)(Dx  ~Rx) But as in earlier examples. No doctors are rich denies another statement. So it could also be treated as a “NOT:” Think about it and hit forward. Problem 7

NOT:some doctor is rich
No doctors are rich. NOT:some doctor is rich So all we have to do is represent some doctor is rich and then put the tilde on it. Hit forward. Problem 7

No doctors are rich. NOT:some doctor is rich Some doctor is rich The statement in red is a standard sort of TISS kind of statement. So we could all turn that into symbols now, I imagine. Hit forward. Problem 7

NOT:some doctor is rich.
No doctors are rich. NOT:some doctor is rich. Some doctor is rich (x)(Dx & Rx) Since this red English statement came from the green statement and the green statement is a “NOT” statement, we now know how to do the green one. Hit forward. Problem 7

No doctors are rich. NOT:some doctor is rich ~(x)(Dx & Rx) Some doctor is rich (x)(Dx & Rx) Since this red English statement is synonymous with the green one, we can see that they should be represented the same way. But remember this was the second way to do this statement. Let us see them both together. Hit forward. Problem 7

NOT:some doctor is rich ~(x)(Dx & Rx)
No doctors are rich. NOT:some doctor is rich ~(x)(Dx & Rx) FET:if it is a doctor then it is not rich. (x)(Dx  ~Rx) Even though these two look quite different, they seem to mean the same thing. They do. We will be able to show their equivalents with future trees or derivations. Hit forward Problem 7

No doctors are rich. NOT:some doctor is rich ~(x)(Dx & Rx) FET:if it is a doctor then it is not rich. (x)(Dx  ~Rx) “No doctors are rich” also allows at least one more interpretation. For if you knew this statement to be true and you found something rich, you could conclude that it was not a doctor Hit forward Problem 7

No doctors are rich. NOT:some doctor is rich ~(x)(Dx & Rx) FET:if it is a doctor then it is not rich. (x)(Dx  ~Rx) FET:if it is rich then it is not a doctor. (x)(Rx  ~Dx) So here is one more way to do this problem. As before, anything logically equivalent to a right answer is also a right answer. So satisfy yourself that these are all logically equivalent before going on to the next problem. Hit forward for next problem. Problem 7

Doctors who are poor are honest. New Problem
This is just slightly more complex than earlier problems. We have 3 properties to deal with.It looks like it would take many things to prove this true but only one to prove it false and so we know it will be a FET kind of statement. So ask yourself if this statement were true, what kind of thing would something have to be in order to conclude that it was something else. For instance, should we say that if we found an honest doctor then he would have to be poor? Think about it and then hit forward. Problem 8

Doctors who are poor are honest.
FET: If it is a poor doctor then it is honest. No, Instead we realize that if this statement were true and we found a poor doctor then we would conclude that he was honest. Thus, that will be the order of the “if then” statement. With our experience we can probably turn this into symbols now. Think about it and then hit forward. Problem 8

FET: If it is a poor doctor then it is honest. (x)((Dx & Px)  Hx) Another way to put it would be to say that if it were a doctor THEN if it were poor then it would be honest. This is an alternative. Think about it and then hit forward. Problem 8

FET: If it is a poor doctor then it is honest. (x)((Dx & Px)  Hx) FET: If it is a doctor then if it is poor then it is honest. (x)(Dx  (Px  Hx)) These are logically equivalent and thus both right answers if one of them is. Hit forward for next problem. Problem 8

Doctors who are poor are non-existent.
New Problem This problems is interesting because it is similar to the last one and yet the dictionary is smaller. The problem asks us to do it with just the concept of “poor” and “doctor”> Existence is not part of the dictionary given and so we must try to represent it by making the quantifiers do the work. So how could we restate this? Think about it and hit forward. Problem 9

It is not the case that there is a poor doctor. This is one way to do it. There are others. For instance if something was a doctor what could you say about it, if the red statement were true? How about if something was poor. Could you conclude anything? Think of these two ways and then hit forward. Problem 9

It is not the case that there is a poor doctor. FET: if it is a doctor then it is not poor. Alternatively FET: if it is poor then it is not a doctor. One of my students wanted to keep the same structure as doctors who are poor are honest and so he started “For Each Thing: if it is a doctor and poor…” How do you think he finished? When ready hit forward. Problem 9

It is not the case that there is a poor doctor. Alternately FET: if it is a doctor then it is not poor. FET: if it is poor then it is not a doctor. FET:if it is a doctor and it is poor then it is not a doctor. O.K. So now let us turn the red and the three greens into symbols. When ready hit forward. Problem 9

It is not the case that there is a poor doctor. ~(x)(Dx & Px) FET: if it is a doctor then it is not poor. (x)(Dx  ~Px) FET: if it is poor then it is not a doctor. (x)(Px  ~Dx) FET:if it is a doctor and it is poor then it is not a doctor. (x)((Px & Dx)  ~Dx) There you have a few of some major ways to represent the statement. But in fact there are an infinite number of correct ways since there are an infinite number of statements that are logically equivalent to these statements. When ready for a new problem hit forward. Problem 9

Think about this and hit forward.
Doctors and lawyers who are rich are admired only if they’re also honest. New Problem There are two problem areas in this statement. One is that we are talking about doctors and lawyers, so, should we start with “For each thing: if it is a Doctor and it is a Lawyer.”? Probably not, for it’s unlikely that the person is talking about people who are both a doctor and a lawyer. One way to handle that would be to start “For each thing: if it is a Doctor OR a Lawyer…” If it is one of those kind of things then we are talking about it. The other problem in this statement is the “only if”. That always causes problems but we know that it is always different from the “if”. The subject is doctors and lawyers who are rich. What should we conclude that if they are honest they are admired or if they are admired then they are honest? Think about this and hit forward. Problem 10

Think about this and hit forward.
Doctors and lawyers who are rich are admired only if they’re also honest. What should we conclude that if they are honest they are admired or if they are admired then they are honest? Notice that if we knew that they were admired then we know the only way that could be is if they were honest. So “honest” should go at the end. Another way to see this is to imagine what it would be like if it were just an “if” instead of an “only if”. Well, it seems that if they were honest, we could conclude that they were admired. And we know that the only if is different. So let us start it as a FET: statement and state the standard form Think about this and hit forward. Problem 10

Doctors and lawyers who are rich are admired only if they’re also honest.
FET: if it is a doctor or a lawyer and it is rich then if it admired then it is honest. Let’s see if we can see the symbols on the next page. Think about this and hit forward. Problem 10

Doctors and lawyers who are rich are admired only if they’re also honest.
FET: if it is a doctor or a lawyer and it is rich then if it admired then it is honest. (x)(((Dx v Lx) & Rx)  (Ax  Hx)) There is another couple of ways to do this statement. One is to put admired on the list of conditions, and only use one horseshoe. The other way would be to make make it an and statement and mirror the and in the original. We could say “All doctors who are rich and admired are honest and All lawyers who are rich and admired are honest .” Imagine how these might look and go forward. Problem 10

a lawyer and it is rich and admired then it is honest
Doctors and lawyers who are rich are admired only if they’re also honest. FET: if it is a doctor or a lawyer and it is rich then if it admired then it is honest. (x)(((Dx v Lx) & Rx)  (Ax  Hx)) FET: if it is a doctor or a lawyer and it is rich and admired then it is honest. (x)((((Dx v Lx) & Rx) & Ax)  Hx) FET: if it is a doctor and it is rich and admired then it is honest and FET: if it is a lawyer and it is rich and admired then it is honest ((x)(((Dx & Rx) & Ax)  Hx) & (x)((( Lx & Rx) & Ax)  Hx)) All these are logically equivalent, as we shall discover later with derivations. Furthermore, they are all synonymous with the green statement. Go forward for new problem Problem 10

All young people are attractive except those who giggle.
What kind of statement is this? Universal (FET) or Existential (TISS). It’s hard to ignore the “All” at the beginning so appears to be a FET statement. The subject of the sentence seems to be young. So we just have to figure out what’s happening with the problem word here, “except” Are we saying that young people who are attractive don’t giggle or young people who don’t giggle are attractive or are we saying both? What would these three look like in standard form? Think about that and go forward with the right arrow. Problem 11

FET: if it’s young and a person and doesn’t giggle then it is attractive. (x)((Yx & Px)  (~Gx  Ax)) You could think of “not giggling” as part of the full condition, so for instance you might think that if someone gives you something young, it is a person and it doesn’t giggle, then you would conclude that it is attractive. So there is that alternative way of approaching it. How would that symbolism look different? Think about that and go forward with the right arrow. Problem 11

FET: if it’s young and a person and doesn’t giggle then it is attractive. (x)((Yx & Px)  (~Gx  Ax)) (x)(((Yx & Px) & ~Gx)  Ax) This works basically because (P  (Q  R)) is equivalent to ((P & Q)  R) When ready for new problem go forward with the right arrow. Problem 11

When ready go forward with the right arrow.
Some of the people on the committee were neither willing nor able to do the job. Clearly the statement in red is an existential statement, since we would only have to find one thing of the right sort to prove it true. What kind of thing would it have to be? Well, it would at least have to be a person on the committee, but what else. Let us create the standard form. When ready go forward with the right arrow. Problem 12

When ready go forward with the right arrow.
Some of the people on the committee were neither willing nor able to do the job. TISS: it is a person and it is on the committee and it’s not willing and not able to do the job. But there is another way. Notice the “nor”. This hints at an ‘or’ way to state that part of the problem. Think about what that might look like. When ready go forward with the right arrow. Problem 12

So let us turn these both into symbols on the next page.
Some of the people on the committee were neither willing nor able to do the job. TISS: it is a person and it is on the committee and it’s not willing and not able to do the job. TISS: it is a person and it is on the committee and it is not willing or able to do the job. So let us turn these both into symbols on the next page. When ready go forward with the right arrow. Problem 12

(Ex)((Px & Cx) & (~Wx & ~Ax))
Some of the people on the committee were neither willing nor able to do the job. TISS: it is a person and it is on the committee and it’s not willing and not able to do the job. (Ex)((Px & Cx) & (~Wx & ~Ax)) TISS: it is a person and it is on the committee and it is not willing or able to do the job. (Ex)((Px & Cx) & ~(Wx v Ax)) Go forward with the right arrow when ready for a new problem. Problem 12

Not all of the books on the shelf are worth reading.
New Problem Consider the statement in red. There are a couple of ways to approach it. One way is to think of it as a “NOT” statement. What does it deny? The other way is to think in terms of what it means. That is what it take to prove it true. .Think about what these standard forms would look like and then hit forward. Problem 13

Alternately TISS: It is a book on the shelf and not worth reading. The first is closest to the structure of English. We are just denying an “All” statement that should pose no problem. It is similar in structure to statements we have already done. But the second is just as good. It notes that to prove the statement true you only need one thing with three specific properties. 1) It is a book. 2) It is on the shelf and 3) it is not worth reading. Write down the representation and then hit forward to verify. Problem 13

~(x)((Bx & Sx)  Wx) TISS: It is a book on the shelf and not worth reading. (x)((Bx & Sx) & ~Wx) This is in contrast to the next statement. Hit forward for next statement. Problem 13

Not one of the books on the shelf is worth reading.
New Problem This also could be approached by considering it as a “NOT”. It is as if someone said “there is a book on the shelf worth reading” and the speaker of the above says “NO!” But consider also that to prove that the speaker is speaking falsely it only takes one thing. A book, on the shelf worth reading. When something can be proved false by one object, that means there is a “FET:” way of stating it. So imagine how we would state these two approaches in standard form. When ready, hit forward. Problem 14

NOT:There is at least one book on the shelf worth reading. Alternately FET:if it is a book on the shelf then it is not worth reading. Imagine how these two red statements could be turned into symbols. When ready, hit forward. Problem 14

NOT:There is at least one book on the shelf worth reading. ~(x)((Bx & Sx) & Wx) FET:if it is a book on the shelf then it is not worth reading. (x)((Bx & Sx)  ~Wx) But there are even other ways to approach it. For we can see that if it’s a book and worth reading then it is not on the shelf and finally something that seems real strange. If it is on the shelf and worth reading then <what do you think?> When ready, hit forward. Problem 14

NOT:There is at least one book on the shelf worth reading. ~(x)((Bx & Sx) & Wx) FET:if it is a book and on the shelf then it is not worth reading. (x)((Bx & Sx)  ~Wx) FET:if it is a book and worth reading then it is not on the shelf. (x)((Bx & Wx)  ~Sx) FET:if it is on the shelf and worth reading then it is not a book. (x)((Sx & Wx)  ~Bx) These are all equivalent and thus all mean the same thing When ready for new problem, hit forward. Problem 14

There are fine universities in Connecticut and Kansas. New Problem
Consider the statement above. There is a common wrong way to do this problem, it is to say that TISS: it is fine it is a university, it is in Connecticut and it is in Kansas Hit forward to see how that would look. Problem 16

There are fine universities in Connecticut and Kansas.
TISS:It is fine, it is a university, it is in Connecticut and it is in Kansas. (x)((Fx & Ux) & (Cx & Kx)) But this can’t be right because what it says is there is a fine university and that university is in both Connecticut and Kansas. Which is possible but not what was intended by the statement. There is usually a student in the class that would like to correct by changing the last “and” to an “or”. Hit forward to see how that would look. Problem 16

TISS:It is fine, it is a university, it is in Connecticut and it is in Kansas. (x)((Fx & Ux) & (Cx & Kx)) TISS:It is fine, it is a university, it is in Connecticut OR it is in Kansas. (x)((Fx & Ux) & (Cx v Kx)) But this can’t be right either since it only says there is at least one. But clearly the speaker wanted there to be at least one in Connecticut and at least one in Kansas. This gives us the clue we need. How would we say that? Hit forward to see it expressed that way. Problem 16

There is a fine university in Connecticut AND there is a fine university in Kansas. The standard for this would be. TISS:It is fine, it is a university, it is in Connecticut AND TISS it is fine, it is a university and it is in Kansas. ((x)((Fx & Ux) & Cx)& (x)((Fx & Ux) & Kx)) So now we see that not all statements involving quantifiers are quantified statements. This is an ampersand statement both sides of which are quantified statements. Hit forward for new problem. Problem 16

Someone can sneak into the ballpark only if George is not on duty.
If someone can sneak into the ballpark then George is not on duty. This is a standard form statement so let us break it up. Imagine the breakup and then hit forward. Problem 17

If someone can sneak into the ballpark then George is not on duty. Someone can sneak into the ballpark George is not on duty. With our experience at this point it is probable that we can now move to their representation. Let P_ be _ is a person, S_: _ can sneak into the ballpark, D_: _is on duty and small ‘g’ for George. Imagine the representation and then hit forward. Problem 17

If someone can sneak into the ballpark then George is not on duty. Someone can sneak into the ballpark George is not on duty. (x)(Px & Sx) ~Dg Because the statement says “someONE”, this implies that the thing is a person, so that’s why we use the ‘P’ there. Why don’t we use a P on the “Duty” side? Because we don’t know that George is a person from the statement. He could be a dog,, for all we know. So how shall we treat the statement in green and thus the statement that it is synonymous to? Imagine the representation and then hit forward. Problem 17

If someone can sneak into the ballpark then George is not on duty. ((x)(Px & Sx)  ~Dg) FET:if it can sneak into the ballpark then George is not on duty. (x)((Px & Sx)  ~Dg) These statements both work, since one works; and they are logically equivalent. In general ((x)Px  Q)is equivalent to (x)(Px  Q) , when the right side if the English statement doesn’t contain the “it” referred to on the left side. But this is not to say that this Existential form is equivalent to a Universal. It is more like a horseshoe statement form has an equivalent universal statement form. Note also that the first red statement does not break our guideline about Existentials not ranging over horseshoes, because the statement form is not ( x)(Px  Q) which is an Existential ranging over a horseshoe it is a horseshoe statement, the left hand side of which is an Existential. For a new problem hit forward. Problem 17

((x)(Px & Sx)  ~Dg) If someone can sneak into the ballpark then George is not on duty. Someone can sneak into the ballpark George is not on duty. (x)(Px & Sx) ~Dg “If then” forms can always be represented with a horseshoe, so that’s why we use it. But we might have approached this problem a little differently. Notice that if anyone sneaks in the we can conclude that George was not on duty. So, this could be a “For each thing:” or “For anything:” kind of statement. Imagine that approach and then hit forward. Problem 17

New Problem Let us address the statement in red. We see that the English logic phrase that is likely to be the most challenging is the “only if”. In fact we can see that the “only if” could be viewed as the major connector, since the left side and right side are both statements when considered independently. So we only need to refer to how the “only if” was treated in logic without quantifiers. When the “only if” is in the middle, it is synonymous to the “If then” standard form with no switching. Imagine then how we will restate this in standard form and then hit forward. Problem 17

Something is missing unless Tom counted incorrectly.
Initially it may look like an Existential statement, since it starts with a “something”. But notice that if we split the statement at the “unless” we have two complete independent statements. So we can think of the “unless” as the major connector and so we just have to figure out what it is equivalent to in standard form. You can use a standard form hint page for this. When ready go forward with the right arrow. Problem 18

Treating the unless as an “or” we can rewrite the it in standard form. When ready go forward with the right arrow. Problem 18

Something is missing or Tom counted incorrectly. Now let’s break it up When ready go forward with the right arrow. Problem 18

Something is missing or Tom counted incorrectly. Something is missing Tom counted incorrectly. We can turn these into symbols now, I think. Think about it and go forward with the right arrow. Problem 18

Something is missing or Tom counted incorrectly. Something is missing Tom counted incorrectly (Ex)Mx ~Ct We can turn these into symbols now, I think. Think about it and go forward with the right arrow. Problem 18

Something is missing or Tom counted incorrectly. Something is missing Tom counted incorrectly (Ex)Mx ~Ct Now, noting that these come from the green statement and it is synonymous with the brown statement. We can finish up Think about it and go forward with the right arrow. Problem 18

Something is missing unless Tom counted incorrectly. ((Ex)Mx  ~Ct)
Something is missing or Tom counted incorrectly. Something is missing Tom counted incorrectly (Ex)Mx ~Ct There you have it Go forward with the right arrow when ready for new problem. Problem 18

If something is worth doing then cynics are wrong.
New Problem Again we can note that between the “if” and the “then” we have a complete and independent statement and after the ‘then’ this statement is complete and independent. Thus the statement in red is in the “if..then” standard form. Let’s break it up. Think about it and go forward with the right arrow. Problem 19

Something is worth doing Cynics are wrong The left side is easy, for the right side we have to decide what is meant. Is it “all cynics are wrong”? Or “some cynics are wrong”? Probably the ‘all’ statement. So we should be able to turn them into symbols now. Think about the symbols and go forward with the right arrow. Problem 19

Something is worth doing Cynics are wrong (x)Dx (x)(Cx  Wx) And since these statements come from the the green one, and that is an “if..then” standard form statement, we can now represent that statement. Think about the symbols and go forward with the right arrow. Problem 19

((x)Dx  (x)(Cx  Wx) Something is worth doing Cynics are wrong (x)Dx (x)(Cx  Wx) So it is a horseshoe statement but it could be expressed another way as “for each thing, if it is worth doing then cynics are wrong” Go forward to see that way. Problem 19

FET:if it is worth doing then cynics are wrong. Let’s pick a specific thing worth doing, like “Algebra Go forward to see the FET fall of and an instance formed. Problem 19

FET:if it is worth doing then cynics are wrong. If Algebra is worth doing then cynics are wrong. This is an if..then statement, so lets break it up. Go forward to see the break up. Problem 19

FET:if it is worth doing then cynics are wrong. If Algebra is worth doing then cynics are wrong. Algebra is worth doing Cynics are wrong We’ve seen how cynics are wrong is represented so let us turn these reds into symbols. Think about the symbols and then go forward. Problem 19

FET:if it is worth doing then cynics are wrong. If Algebra is worth doing then cynics are wrong. Algebra is worth doing Cynics are wrong Da (x)(Cx  Wx) Let’s do the green, which is an “if..then” statement Think about the symbols and then go forward. Problem 19

FET:if it is worth doing then cynics are wrong. If Algebra is worth doing then cynics are wrong. (Da  (x)(Cx  Wx)) Algebra is worth doing Cynics are wrong Da (x)(Cx  Wx) The red is an instance of the green, so how will we represent the green? Think about the symbols and then go forward. Problem 19

FET:if it is worth doing then cynics are wrong. (y)(Dy  (x)(Cx  Wx)) If Algebra is worth doing then cynics are wrong. (Da  (x)(Cx  Wx)) Algebra is worth doing Cynics are wrong Da (x)(Cx  Wx) We could not pick ‘x’ for the first quantifier in the red formal statement because that would have been a confused statement. We wouldn’t know whether the ‘x’ in “Cx’ referred to the first quantifier or the second. It’s not a legal formal statement if one quantifier overlaps another with the same variable So let’s put the two representations on the same page. Go forward. Problem 19

((x)Dx  (x)(Cx  Wx) FET:if it is worth doing then cynics are wrong. (y)(Dy  (x)(Cx  Wx)) These statements are synonymous and thus logically equivalent. Again we see a horseshoe statement with an existential on the left equivalent to a universal statement. Go forward for new problem. Problem 19

If something is worth doing then it’s worth doing well.
New Problem This statement looks like it is in the same form as the last one, but not quite. Notice “Something is worth doing” is a complete and independent statement but “It’s worth doing well” is not independent of what was being talked about on the left side. The “IT’ on the right side refers back to the thing worth doing. So that means that this is a quantified statement and it is clear that a horseshoe goes in the middle. But we know that existentials do not overlap horseshoes as the next major connector. So this must be a universal statement. We should start it with “FET” Think about the standard form and go forward when you need confirmation. Problem 20

If something is worth doing then it’s worth doing well.
FET:if it is worth doing then it is worth doing well. This was much simpler than we suspected. We should be able to represent the red and the green in one more page. Think about the representation and go forward. Problem 20

If something is worth doing then it’s worth doing well. (x)(Dx  Wx)
FET:if it is worth doing then it is worth doing well. This was much simpler than we suspected. We should be able to represent the red and the green in one more page. Go forward for new problem. Problem 20

If all tax evaders are caught then uncle Joe will be. New Problem
Should we think of this statement as a universal or existential? Answer and go forward. Problem 21

If all tax evaders are caught then uncle Joe will be.
Should we think of this statement as a universal or existential? Neither. It is an “if..then” statement. Notice that “All tax evaders are caught” is a statement and with a little completion “Uncle Joe will be caught” is also a statement. So let us treat is as an “if..then” standard form and break it up. Imagine and go forward. Problem 21

All tax evaders are caught Uncle Joe will be caught These are now very straightforward statements. Let us represent them with T_, C_ and j for Uncle Joe, Imagine and go forward. Problem 21

All tax evaders are caught Uncle Joe will be caught (x)(Tx  Cx) Cj So now representing the green statement will be no problem. Imagine and go forward. Problem 21

((x)(Tx  Cx)  Cj) All tax evaders are caught Uncle Joe will be caught (x)(Tx  Cx) Cj . Go forward for new problem. Problem 21

If any tax evader is caught then uncle Joe will be. New Problem
This is different. Who would have thought there would be a difference between all and any? We can’t really break this down into an “if..then” statement without a bit of modification, because “Any tax evader is caught” is not really a statement. But there are a couple of ways to do it. One is to see that this says that “If some tax evader is caught then uncle Joe will be caught” This is in standard form. Think and then go forward. Problem 22

If any tax evader is caught then uncle Joe will be.
If some tax evader is caught then uncle Joe will be caught Some tax evader is caught Uncle Joe will be caught. So now we can represent the left and represent the right. Think what that will look like, and then go forward. Problem 22

If some tax evader is caught then uncle Joe will be caught Some tax evader is caught Uncle Joe will be caught. (x)(Tx & Cx) Cj Now we can see how to represent the standard form green statement and the brown statement that it is synonymous with. Think what that will look like, and then go forward. Problem 22

((x)(Tx & Cx)  Cj) If some tax evader is caught then uncle Joe will be caught Some tax evader is caught Uncle Joe will be caught. (x)(Tx & Cx) Cj The other way to do it is to realize that what this says is for anything: if it is a tax evader and caught, then Uncle Joe will be caught, A FET statement. Think what that will look like, and then go forward. Problem 22

FET:If it is a tax evader and caught then uncle Joe will be caught. Taking as an instance. “ If Sam is a tax evader and caught then uncle Joe will be caught.” Think what that will look like, and then go forward. Problem 22

FET:If it is a tax evader and caught then uncle Joe will be caught. If Sam is a tax evader and caught then uncle Joe will be caught. This should be an easy if..then” statement. Let us represent it. Think what that will look like, and then go forward. Problem 22

FET:If it is a tax evader and caught then uncle Joe will be caught. If Sam is a tax evader and caught then uncle Joe will be caught. ((Ts & Cs)  Cj) But Sam was just an instance or example of the ‘FET’ statement, so let us represent the ‘FET’ statement by taking Sam out and putting x in. Think what that will look like, and then go forward. Problem 22

FET:If it is a tax evader and caught then uncle Joe will be caught. (x)((Tx & Cx)  Cj) If Sam is a tax evader and caught then uncle Joe will be caught. ((Ts & Cs)  Cj) So this is another way to represent the green and so we see again that a horseshoe statement where the left hand side is an existential can be expressed as a universal. Lets see them together Think what that will look like, and then go forward. Problem 22

(x)((Tx & Cx)  Cj) FET:If it is a tax evader and caught then uncle Joe will be caught. If some tax evader is caught then uncle Joe will be caught ((x)(Tx & Cx)  Cj) Notice that we cannot infer from the green or either representation that Joe is a tax evader. He could just be the dope smoker in the house where all the tax evaders are, for we know, he might be caught in the raid. Go forward for new problem Problem 22

If someone makes a joke and no one laughs, then she feels silly.
New Problem Notice again the “she feels silly” is dependent for its meaning on the person who made a joke. Thus, this is not a horseshoe statement but clearly it has a horseshoe in it. So it must be a “FET” statement. Go forward for standard form. Problem 23

FET:if it is a person and it makes a joke and no one laughs then it feels silly. It’s a “FET” statement so let us put in an instance, like Robin Williams Go forward for instance. Problem 23

FET:if it is a person and it makes a joke and no one laughs then it feels silly. IF Bush is a person and Bush makes a joke and no one laughs then Bush feels silly. Let’s break up this if then statement Imagine and go forward when ready. Problem 23

FET:if it is a person and it makes a joke and no one laughs then it feels silly. IF Bush is a person and Bush makes a joke and no one laughs then Bush feels silly. Bush is a person and Bush makes a joke and no one laughs Bush feels silly. We could have had the dictionary where to express “makes a joke” we had to say that there is at least one joke that Bush makes it. But let us instead simplify the dictionary to be “x makes a joke” . But let “x laughs” be part of the dictionary. Let’s break down the left side a little more before symbolizing. Imagine and go forward when ready. Problem 23

FET:if it is a person and it makes a joke and no one laughs then it feels silly. IF Bush is a person and Bush makes a joke and no one laughs then Bush feels silly. Bush is a person and Bush makes a joke and no one laughs Bush feels silly. Sb Bush is a person and Bush makes a joke No one laughs No one laughs should be “FET: if it’s a person then it does not laugh” or “It is not the case that there is something that is a person and laughs.” Let’s pick the first and symbolize. Imagine and go forward when ready. Problem 23

FET:if it is a person and it makes a joke and no one laughs then it feels silly. IF Bush is a person and Bush makes a joke and no one laughs then Bush feels silly. Bush is a person and Bush makes a joke and no one laughs Bush feels silly. ((Pb & Jb) & (x)(Px  ~Lx)) Sb Bush is a person and Bush makes a joke No one laughs (Pb & Jb) (x)(Px  ~Lx) Let’s do the green. Imagine and go forward when ready. Problem 23

FET:if it is a person and it makes a joke and no one laughs then it feels silly. IF Bush is a person and Bush makes a joke and no one laughs then Bush feels silly. (((Pb & Jb) & (x)(Px  ~Lx))  Sb) Bush is a person and Bush makes a joke and no one laughs Bush feels silly. ((Pb & Jb) & (x)(Px  ~Lx)) Sb Bush is a person and Bush makes a joke No one laughs (Pb & Jb) (x)(Px  ~Lx) Let’s do the green, the red is an instance of it and we have to pick a new variable. Imagine and go forward when ready. Problem 23

FET:if it is a person and it makes a joke and no one laughs then it feels silly. (y)(((Py & Jy) & (x)(Px  ~Lx))  Sy) IF Bush is a person and Bush makes a joke and no one laughs then Bush feels silly. (((Pb & Jb) & (x)(Px  ~Lx))  Sb) Bush is a person and Bush makes a joke and no one laughs Bush feels silly. ((Pb & Jb) & (x)(Px  ~Lx)) Sb Bush is a person and Bush makes a joke No one laughs (Pb & Jb) (x)(Px  ~Lx) Let’s do the green. It is synonymous with the red. Imagine and go forward when ready. Problem 23

(y)(((Py & Jy) & (x)(Px  ~Lx))  Sy) FET:if it is a person and it makes a joke and no one laughs then it feels silly. IF Bush is a person and Bush makes a joke and no one laughs then Bush feels silly. (((Pb & Jb) & (x)(Px  ~Lx))  Sb) Bush is a person and Bush makes a joke and no one laughs Bush feels silly. ((Pb & Jb) & (x)(Px  ~Lx)) Sb Bush is a person and Bush makes a joke No one laughs (Pb & Jb) (x)(Px  ~Lx) So there you have it. A somewhat difficult problem Go forward when ready for new problem. Problem 23

Go forward to see development.
If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. New Problem With this statement, it would be tempting to think of this as an existential. But then we realize that in the middle of it is a horseshoe and on the right side there is another reference to the same person “she”. So we get the hint that really this is about anyone who tries hard. The form “If some.” is often a universal. So let us start this as a “For Each Thing:” statement. Go forward to see development. Problem 24

If someone tries hard, then she’ll succeed unless some people who try hard have bad luck.
FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. Let us name the thing Alan and see how it looks Go forward to see development. Problem 24

Now it’s time to break it up
If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. Now it’s time to break it up Imagine and go forward Problem 24

If someone tries hard, then she’ll succeed unless some people who try hard have bad luck.
FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. Alan is a person and Alan tries hard Alan will succeed unless some people who try hard have bad luck. The left side is easy (Pa&Ha), but how about the right side? It is an unless statement which is synonymous with an ‘OR’ standard form statement. So let us restate the right side Imagine and go forward Problem 24

(Pa & Ha) try hard have bad luck.
If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. Alan is a person and Alan tries hard Alan will succeed unless some people who (Pa & Ha) try hard have bad luck. Alan will succeed or some people who try hard have bad luck. Now the red statement is an “OR’ standard form statement. Let’s break it up Imagine and go forward Problem 24

If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. Alan is a person and Alan tries hard Alan will succeed unless some people who (Pa & Ha) try hard have bad luck. Alan will succeed or some people who try hard have bad luck. Alan will succeed Some people who try hard have bad luck. Alan will succeed is simply “Sa”. “Some people who try hard have bad luck” is clearly an existential. Imagine what single thing you could find to prove this statement with certainty. Imagine and go forward Problem 24

If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. Alan is a person and Alan tries hard Alan will succeed unless some people who (Pa & Ha) try hard have bad luck. Alan will succeed or some people who try hard have bad luck. Alan will succeed Some people who try hard have bad luck Sa TISS: it is a person, it tries hard and it has bad luck. So now we should be able to turn the red statement into symbols. Imagine and go forward Problem 24

If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. Alan is a person and Alan tries hard Alan will succeed unless some people who (Pa & Ha) try hard have bad luck. Alan will succeed or some people who try hard have bad luck. Alan will succeed Some people who try hard have bad luck Sa (x)((Px & Hx) & Bx) . TISS: it is a person, it tries hard and it has bad luck. (x)((Px & Hx) & Bx) These two statements are from an “OR” statement, so let’s form that one. Go forward. Problem 24

If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. Alan is a person and Alan tries hard Alan will succeed unless some people who (Pa & Ha) try hard have bad luck. Alan will succeed or some people who try hard have bad luck. (Sa v (x)((Px & Hx) & Bx) . Alan will succeed Some people who try hard have bad luck Sa (x)((Px & Hx) & Bx) . TISS: it is a person, it tries hard and it has bad luck. (x)((Px & Hx) & Bx) The red is synonymous with the green so we can put in that representation. Go forward. Problem 24

((Pa & Ha)  (Sa v (x)((Px & Hx) & Bx))
If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. ((Pa & Ha)  (Sa v (x)((Px & Hx) & Bx)) Alan is a person and Alan tries hard Alan will succeed unless some people who (Pa & Ha) try hard have bad luck. (Sa v (x)((Px & Hx) & Bx) Alan will succeed or some people who try hard have bad luck. (Sa v (x)((Px & Hx) & Bx) . Alan will succeed Some people who try hard have bad luck Sa (x)((Px & Hx) & Bx) . TISS: it is a person, it tries hard and it has bad luck. (x)((Px & Hx) & Bx) The red statement is an instance of the green statement so let us change “Alan” to y and put on a quantifier. Go forward. Problem 24

(x)((Py & Hy)  (Sy v (x)((Px & Hx) & Bx))
If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. (x)((Py & Hy)  (Sy v (x)((Px & Hx) & Bx)) If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. ((Pa & Ha)  (Sa v (x)((Px & Hx) & Bx)) Alan is a person and Alan tries hard Alan will succeed unless some people who (Pa & Ha) try hard have bad luck. (Sa v (x)((Px & Hx) & Bx) Alan will succeed or some people who try hard have bad luck. (Sa v (x)((Px & Hx) & Bx) . Alan will succeed Some people who try hard have bad luck Sa (x)((Px & Hx) & Bx) . TISS: it is a person, it tries hard and it has bad luck. (x)((Px & Hx) & Bx) The green statement is synonymous, so it will be represented the same way. Go forward. Problem 24

(y)((Py & Hy)  (Sy v (x)((Px & Hx) & Bx))
If someone tries hard, then she’ll succeed unless some people who try hard have bad luck. (y)((Py & Hy)  (Sy v (x)((Px & Hx) & Bx)) FET:If it is a person and it tries hard, then it will succeed unless some people who try hard have bad luck. If Alan is a person and Alan tries hard, then Alan will succeed unless some people who try hard have bad luck. ((Pa & Ha)  (Sa v (x)((Px & Hx) & Bx)) Alan is a person and Alan tries hard Alan will succeed unless some people who (Pa & Ha) try hard have bad luck. (Sa v (x)((Px & Hx) & Bx) Alan will succeed or some people who try hard have bad luck. (Sa v (x)((Px & Hx) & Bx) . Alan will succeed Some people who try hard have bad luck Sa (x)((Px & Hx) & Bx) . TISS: it is a person, it tries hard and it has bad luck. (x)((Px & Hx) & Bx) There we have it. Go forward for new problem. Problem 24

All draft dodgers who have gone to Canada will be free from prosecution if any one of them is.
New Problem What does this mean? Try to state it in a standard form and then go forward to see if it is the same as mine. Problem 27

All draft dodgers who have gone to Canada will be free from prosecution if any one of them is.
If some draft dodger who goes to Canada is free from prosecution then all draft dodgers who goes to Canada will be free from prosecution This is an ‘if..then” standard form statement so we can represent it when we represent its parts, then just put a horseshoe in the center. Go forward when ready Problem 27

It shouldn’t be very hard to turn these into symbols Go forward
All draft dodgers who have gone to Canada will be free from prosecution if any one of them is. If some draft dodger who goes to Canada is free from prosecution then all draft dodgers who goes to Canada will be free from prosecution Some draft dodger who goes to Canada is All draft dodgers who goes to free from prosecution Canada will be free from prosecution. It shouldn’t be very hard to turn these into symbols Go forward Problem 27

Now we can build the green statement Go forward
All draft dodgers who have gone to Canada will be free from prosecution if any one of them is. If some draft dodger who goes to Canada is free from prosecution then all draft dodgers who goes to Canada will be free from prosecution Some draft dodger who goes to Canada is All draft dodgers who goes to free from prosecution Canada will be free from prosecution. (x)(Dx &(Cx & Fx)) (x)((Dx &Cx)  Fx) Now we can build the green statement Go forward Problem 27

((x)(Dx &(Cx & Fx))  (x)((Dx &Cx)  Fx))
All draft dodgers who have gone to Canada will be free from prosecution if any one of them is. If some draft dodger who goes to Canada is free from prosecution then all draft dodgers who goes to Canada will be free from prosecution ((x)(Dx &(Cx & Fx))  (x)((Dx &Cx)  Fx)) Some draft dodger who goes to Canada is All draft dodgers who goes to free from prosecution Canada will be free from prosecution. (x)(Dx &(Cx & Fx)) (x)((Dx &Cx)  Fx) And since the green is equivalent, we can represent it now. Go forward Problem 27

((x)(Dx &(Cx & Fx))  (x)((Dx &Cx)  Fx))
All draft dodgers who have gone to Canada will be free from prosecution if any one of them is. ((x)(Dx &(Cx & Fx))  (x)((Dx &Cx)  Fx)) If some draft dodger who goes to Canada is free from prosecution then all draft dodgers who goes to Canada will be free from prosecution Some draft dodger who goes to Canada is All draft dodgers who goes to free from prosecution Canada will be free from prosecution. (x)(Dx &(Cx & Fx)) (x)((Dx &Cx)  Fx) When ready for new problem, go forward Problem 27

Bob likes anyone who pays attention to him.
New Problem What does this red statement mean? Well if x pays attention to Bob then what does this statement say is also true about x? Think about it and then go forward Problem 31

FET:if it is a person and it pays attention to Bob, then Bob likes it. The fact that the statement says “anyONE” is an indicator that we are talking about any person who pays attention to Bob. So if someone pays attention to Bob, then Bob likes that person. So now let’s plug in an example person. Think about it and then go forward Problem 31

FET:if it is a person and it pays attention to Bob, then Bob likes it. If Ann is a person and Ann pays attention to Bob, then Bob likes Ann. Now it’s in standard form so let’s break it up. Think about it and then go forward Problem 31

FET:if it is a person and it pays attention to Bob, then Bob likes it. If Ann is a person and Ann pays attention to Bob, then Bob likes Ann. Ann is a person and Ann pays attention to Bob Bob likes Ann. Let us suppose the dictionary is Axy:x pays attention to y and Lxy: x likes y, and of course Px: x is a person. Let’s represent these red statements. Think about it and then go forward Problem 31

FET:if it is a person and it pays attention to Bob, then Bob likes it. If Ann is a person and Ann pays attention to Bob, then Bob likes Ann. Ann is a person and Ann pays attention to Bob Bob likes Ann. (Pa & Aab) Lba Notice that the order of terms is switched. ‘a’ pays attention to ‘b’ and ‘b’ likes ‘a’. So let us build the green now. Think about it and then go forward Problem 31

FET:if it is a person and it pays attention to Bob, then Bob likes it. If Ann is a person and Ann pays attention to Bob, then Bob likes Ann. ((Pa & Aab)  Lba) Ann is a person and Ann pays attention to Bob Bob likes Ann. (Pa & Aab) Lba “Ann” is an instance of the FET statement, so let’s get the green now. Think about it and then go forward Problem 31

FET:if it is a person and it pays attention to Bob, then Bob likes it. (x)((Px & Axb)  Lbx) If Ann is a person and Ann pays attention to Bob, then Bob likes Ann. ((Pa & Aab)  Lba) Ann is a person and Ann pays attention to Bob Bob likes Ann. (Pa & Aab) Lba And of course the red is synonymous with the green. Think about it and then go forward Problem 31

Bob likes anyone who pays attention to him. (x)((Px & Axb)  Lbx)
FET:if it is a person and it pays attention to Bob, then Bob likes it. If Ann is a person and Ann pays attention to Bob, then Bob likes Ann. ((Pa & Aab)  Lba) Ann is a person and Ann pays attention to Bob Bob likes Ann. (Pa & Aab) Lba There you have it. Go forward when ready for a new problem. Problem 31

If Tom doesn't like anyone, then he surely doesn't like everyone.
New Problem Again we have an “if..then” statement since the left and right side are complete and independent statements. Not quite though. Since “he” on the right refers to Tom. But we could rewrite it easily as an “if..then” statement with that substitution. So let’s imagine that is done and break it up on the next slide. Go forward when ready. Problem 32

If Tom doesn't like anyone, then he surely doesn't like everyone
Tom doesn't like anyone Tom doesn't like everyone. The “surely” was not part of the content of the statement but reflected instead something about the attitude of the speaker. There are two natural ways to symbolized both of these statements. Imagine each and then go when ready. Problem 32

Tom doesn't like anyone Tom doesn't like everyone. ~(x)(Px & Ltx) ~(x)(Px  Lx) (x)(Px  ~Ltx) (x)(Px & ~Lx) The brown statements are probably closest to the structure of the English but anything logically equivalent to a right answer is also a right answer so the red formal statements also work. They are more like what you would immediately infer from the English red statements. Now we see the red English statements come from the green, so let’s put the green together. Hit forward when ready. Problem 32

(~(x)(Px & Ltx)  ~(x)(Px  Lx)) (~(x)(Px & Ltx)  (x)(Px & ~Lx)) ((x)(Px  ~Ltx)  ~(x)(Px  Lx)) ((x)(Px  ~Ltx)  (x)(Px & ~Lx)) Tom doesn't like anyone Tom doesn't like everyone. ~(x)(Px & Ltx) ~(x)(Px  Lx) (x)(Px  ~Ltx) (x)(Px & ~Lx) There are 4 ways to put together the target statement. All are logically equivalent. Unfortunately equivalence does not always mirror the complex structure of a statement, since all of these statements are tantamount to just claiming that there is at least one person in the universe. (x)Px. Now I guess we see why the statement said “surely”. Hit forward when ready for a new problem. Problem 32

There’s no time like the present
New Problem Well, what does this mean? We can approach it two different ways,, both of which are fairly natural. The first is to consider what doesn’t exist according to the statement. Think about that and then go forward. Problem 33

There’s no time like the present.
NOT: There is exists a time like the present time A “NOT” statement. So let us worry about the statement without the not. Think about that and then go forward. Problem 33

NOT: There is exists a time like the present time. There is exists a time like the present time. We should be able to turn this into symbols on the next slide., using Tx:x is a time, Lxy:x is like y and p: the present time Think about that and then go forward. Problem 33

NOT: There is exists a time like the present time. There is exists a time like the present time. (x)(Tx & Lxp) So, the green and the brown should be easy to figure. Think about that and then go forward. Problem 33

There’s no time like the present. ~(x)(Tx & Lxp)
NOT: There is exists a time like the present time. There is exists a time like the present time. (x)(Tx & Lxp) But we know that whenever we have a tilde existential, we could also do it as a universal. So how would that look? Think about the symbols and then go forward. Problem 33

There’s no time like the present. ~(x)(Tx & Lxp) (x)(Tx  ~Lxp)
For any thing:if it is a time then it is not like the present time. But we know that the present time itself is a time. So what does this statement say about the present time? That it is not even like itself! This can’t be But if we wanted to change the statement to something that could at least be true then we could say that there is no time other than the present like the present. We might need identity for this. Let us use the universal form and try to represent that. Think about the symbols and then go forward. Problem 33

(x)((Tx & ~x=p) ~Lxp) Alternately ~(x)((Tx & ~x=p) & Lxp)
There’s no time like the present, other than the present. ~(x)(Tx & Lxp) (x)((Tx & ~x=p) ~Lxp) Alternately ~(x)((Tx & ~x=p) & Lxp) This identity connector will be introduced in this section, since it has a variety of uses and comes up a lot. Go forward when ready for a new problem. Problem 33

Tom is taller than anyone he knows.
New Problem This is clearly a FET statement, but which way should the horseshoe go? Is it if Tom is taller than someone then he knows him, or the other way around? Think about this and then go forward. Problem 34

FET:if it is a person and Tom knows it then Tom is taller than it We should be able to turn this into symbols now. Think about this and then go forward. Problem 34

FET:if it is a person and Tom knows it then Tom is taller than it. (x)((Px & Ktx)  Ttx) And of course the green statement should be symbolized the same way. Go forward for new problem. Problem 34

Tom has confidence in himself only if someone else does also.
New Problem The only if divides two independent statements, almost independent anyway. We have to rewrite the second part to read someone else has confidence in Tom. The only if we have seen before and according to our translation hints should just be an “if..then” without switching. So let’s restate it in standard form Go forward when ready. Problem 35

If Tom has confidence in himself then someone else has confidence in Tom. Let’s break it up. Go forward when ready. Problem 35

If Tom has confidence in himself then someone else has confidence in Tom. Tom has confidence in himself. Someone else has confidence in Tom. Tom has confidence in himself is easy and atomic ‘Ctt’ But someone ELSE has confidence in Tom means someone other than Tom has confidence in Tom. Let’s try symbolizing that next page. Go forward when ready. Problem 35

If Tom has confidence in himself then someone else has confidence in Tom. Tom has confidence in himself. Someone else has confidence in Tom. Ctt (x)((Px & ~x=t) & Cxt) To say that someone else has confidence in Tom means that you have a person that is not identical to Tom but has confidence in Tom. So now we should be able to represent the green and therefore the brown statements. Go forward when ready. Problem 35

(Ctt  (x)((Px & ~x=t) & Cxt)) If Tom has confidence in himself then someone else has confidence in Tom. Tom has confidence in himself. Someone else has confidence in Tom. Ctt (x)((Px & ~x=t) & Cxt) But wait a minute. Maybe we misinterpreted the statement. Maybe what it means is that Tom has confidence in himself only if someone else has confidence in themselves. How would that look different? Go forward when ready. Problem 35

If Tom has confidence in himself then someone else has confidence in Tom. (Ctt  (x)((Px & ~x=t) & Cxt)) If Tom has confidence in himself then someone else has confidence in themselves. (Ctt  (x)((Px & ~x=t) & Cxx)) These are not equivalent but depend upon our interpretation of the original statement Go forward for new problem. Problem 35

Everybody doesn’t like something, but nobody doesn’t like Sara Lee
New Problem This looks like a ‘but’ statement, which is synonymous with an ‘and’ form. So let’s skip a step and break it into its parts Go forward when ready. Problem 36

Everybody doesn’t like something. Nobody doesn’t like Sara Lee. The left side means what? That everyone has something that they don’t like. The second part has a couple of negatives in it. We might try restating it more positively, but let’s not. Let us try to capture all of the tildes involved. Symbols for these guys next step. Go forward when ready. Problem 36

Everybody doesn’t like something. Nobody doesn’t like Sara Lee. (x)(Px  (y)~Lxy) ~(x)(Px & ~Lxs) The second says. It is not the case that there exists something such that it is a person and doesn’t like Sara Lee. So we can now do the green statement Go forward when ready. Problem 36

((x)(Px  (y)~Lxy) & ~(x)(Px & ~Lxs))
Everybody doesn’t like something, but nobody doesn’t like Sara Lee ((x)(Px  (y)~Lxy) & ~(x)(Px & ~Lxs)) Everybody doesn’t like something. Nobody doesn’t like Sara Lee. (x)(Px  (y)~Lxy) ~(x)(Px & ~Lxs) Also we can see that “nobody doesn’t like Sara Lee”, just means “everyone likes Sara Lee”. So we could have done this without negations on the right side. Go forward when ready. Problem 36

((x)(Px  (y)~Lxy) & (x)(Px  Lxs)) Everybody doesn’t like something. Nobody doesn’t like Sara Lee. (x)(Px  (y)~Lxy) (x)(Px  Lxs) So that’s an alternative. Go forward when ready for a new problem. Problem 36

Go forward when ready for split.
Everybody likes, and is liked by, somebody; but nobody likes, or is liked by, everybody. New Problem This is fairly involved statement. It is a “but” statement which is synonymous with an “and” form but each side is relatively involved. So let’s split up and then deal with each part separately. Go forward when ready for split. Problem 37

Everybody likes, and is liked by, somebody; but nobody likes, or is liked by, everybody. Everybody likes, and is liked Nobody likes, or is liked by by, somebody. Everybody. (37a) (37b) Let’s split this problem 37 into 37a, the left side and 37b, the right side. Now let’s go deal with problem 37a on the next slide. Go forward when ready for split. Problem 37

Everybody likes, and is liked by, somebody.
But what does this say. In particular does it say that everybody likes somebody and everybody is liked by somebody and maybe they are different people? Or does it say that everybody has somebody that they like and are liked by? We may be able to tell once we break down the form. For now let us state it as a FET statement changing as little as possible. Go forward when ready for standard form FET statement. Problem 37a

FET: if it is a person then it likes, and is liked by, somebody. Now let’s plug in a likeable person,like Janine Garofolo maybe. Go forward when ready for plug in. Problem 37a

FET: if it is a person then it likes, and is liked by, somebody. If Janine is a person then Janine likes, and is liked, by somebody. As an “if..then” standard form, we can split it up Go forward when ready for split up. Problem 37a

FET: if it is a person then it likes, and is liked by, somebody. If Janine is a person then Janine likes, and is liked, by somebody. Janine is a person Janine likes, and is liked, by somebody. “Janine is a person” is easy and atomic. But now we see that “Janine likes and is liked by somebody” is more likely referring to the same person. So let’s represent the left and restate the right in a standard form. Go forward when ready . Problem 37a

FET: if it is a person then it likes, and is liked by, somebody. If Janine is a person then Janine likes, and is liked, by somebody. Janine is a person Janine likes, and is liked, by somebody. Pj TISS:it is a person and Janine likes and is liked by it. Go forward when ready . Problem 37a

FET: if it is a person then it likes, and is liked by, somebody. If Janine is a person then Janine likes, and is liked, by somebody. Janine is a person Janine likes, and is liked, by somebody. Pj TISS:it is a person and Janine likes and is liked by it. So at this point, with the experience that I assume we have, we can turn the TISS statement into symbols. Lets try that. Go forward when ready . Problem 37a

FET: if it is a person then it likes, and is liked by, somebody. If Janine is a person then Janine likes, and is liked, by somebody. Janine is a person Janine likes, and is liked, by somebody. Pj TISS:it is a person and Janine likes and is liked by it. (x)((Px & Lxj) & Ljx) The green statement is represented the same as the “TISS” statement. The brown is a simple “if..then” statement. Knowing that we can build the green and brown in one slide, I hope. Think first before forward. Go forward when ready . Problem 37a

FET: if it is a person then it likes, and is liked by, somebody. If Janine is a person then Janine likes, and is liked, by somebody. (Pj  (x)((Px & Lxj) & Ljx)) Janine is a person Janine likes, and is liked, by somebody. Pj (x)((Px & Lxj) & Ljx) TISS:it is a person and Janine likes and is liked by it. (x)((Px & Lxj) & Ljx) The green statement if a FET for which Janine was an example and the brown statement means the same thing. So let us build those in one slide. Think about it and go forward when ready . Problem 37a

(x)(Py  (x)((Px & Lxy) & Lyx)) FET: if it is a person then it likes, and is liked by, somebody. If Janine is a person then Janine likes, and is liked, by somebody. (Pj  (x)((Px & Lxj) & Ljx)) Janine is a person Janine likes, and is liked, by somebody. Pj (x)((Px & Lxj) & Ljx) TISS:it is a person and Janine likes and is liked by it. (x)((Px & Lxj) & Ljx) We had to pick a something other than “x” here to stand for Janine, the any person, because the quantifier overlaps the existential x quantifier, so we chose “y”. Let’s remember though that this problem is just 37a, which is half of 37. We now turn our attention to 37b, the other half of the original “but” statement. Think about it and go forward to the other half when ready . Problem 37a

Nobody likes, or is liked by everybody.
How shall we do this? Should we say that “It is not the case that there exists”? Or should we say “for each thing…” Both are acceptable ways to start. This may be a personal preference as to which would be more intuitive. Let’s take the first approach, and consider the other later. It is not the case that. Think about it and go forward when ready . Problem 37b

Nobody likes, or is liked by, everybody.
NOT:TISS: it is likes, or is liked by, everybody. Let’s do two steps to the next page. Peel the “NOT” and plug something in for the TISS. Think about it and go forward when ready . Problem 37b

NOT:TISS: it likes, or is liked by, everybody. TISS: it likes, or is liked by, everybody. Janine likes, or is liked by, everybody It might have seemed ambiguous before we got to Janine, and it may still seem that way. Does this mean that anyone you pick Janine either likes them or they like her? So the first person she likes but the next person likes her, etc? Or does it mean either she likes everybody or everybody likes her. It seems more like the latter to me, so let’s go with that Think about it and go forward when ready . Problem 37b

NOT:TISS: it likes, or is liked by, everybody. TISS: it likes, or is liked by, everybody. Janine likes, or is liked by, everybody Janine likes everybody or Janine is liked by everybody Let’s split it up. Think about it and go forward when ready . Problem 37b

NOT:TISS: it likes, or is liked by, everybody. TISS: it likes, or is liked by, everybody. Janine likes, or is liked by, everybody Janine likes everybody or Janine is liked by everybody Janine likes everybody Janine is liked by everybody. Let me hope we can turn each of these into formal statements in the next slide. Think about it and go forward when ready . Problem 37b

NOT:TISS: it likes, or is liked by, everybody. TISS: it likes, or is liked by, everybody. Janine likes, or is liked by, everybody Janine likes everybody or Janine is liked by everybody Janine likes everybody Janine is liked by everybody. (x)(Px  Ljx) (x)(Px  Lxj) Let’s build the green and the brown in one slide, since the green is just a wedge fof these two reds and the brown is synonymous. Think about it and go forward when ready . Problem 37b

NOT:TISS: it likes, or is liked by, everybody. TISS: it likes, or is liked by, everybody. Janine likes, or is liked by, everybody ((x)(Px  Ljx) v (x)(Px  Lxj)) Janine likes everybody or Janine is liked by everybody Janine likes everybody Janine is liked by everybody. (x)(Px  Ljx) (x)(Px  Lxj) Let’s build the green and the brown in one slide, since the green is just a wedge fof these two reds and the brown is synonymous. Think about it and go forward when ready . Problem 37b

NOT:TISS: it likes, or is liked by, everybody. TISS: it likes, or is liked by, everybody. Janine likes, or is liked by, everybody ((x)(Px  Ljx) v (x)(Px  Lxj)) Janine likes everybody or Janine is liked by everybody Janine likes everybody Janine is liked by everybody. (x)(Px  Ljx) (x)(Px  Lxj) Janine was just an example for the TISS statement, so let’s try the green now. Think about it and go forward when ready . Problem 37b

NOT:TISS: it likes, or is liked by, everybody. TISS: it likes, or is liked by, everybody. (y)((x)(Px  Lyx) v (x)(Px  Lxy)) Janine likes, or is liked by, everybody ((x)(Px  Ljx) v (x)(Px  Lxj)) Janine likes everybody or Janine is liked by everybody Janine likes everybody Janine is liked by everybody. (x)(Px  Ljx) (x)(Px  Lxj) The red is just the green without a “NOT” and the green is synonymous with the brown, so let’s do both of those in one slide. Think about it and go forward when ready . Problem 37b

~(y)((x)(Px  Lyx) v (x)(Px  Lxy)) NOT:TISS: it likes, or is liked by, everybody. TISS: it likes, or is liked by, everybody. (y)((x)(Px  Lyx) v (x)(Px  Lxy)) Janine likes, or is liked by, everybody ((x)(Px  Ljx) v (x)(Px  Lxj)) Janine likes everybody or Janine is liked by everybody Janine likes everybody Janine is liked by everybody. (x)(Px  Ljx) (x)(Px  Lxj) If you push the tilde all the way through the statement, you would get that every person has someone the don’t like and that doesn’t like them. It would look like this: ( y)((x)(Px & ~Lyx) & (x)(Px & ~Lxy)) And if you recall, this is just problem 37b, half of 37, so let’s put 37a & 37b together to see the final answer. Think about it and go forward when ready . Problem 37b

Everybody likes, and is liked by, somebody; but nobody likes, or is liked by, everybody. Everybody likes, and is liked Nobody likes, or is liked by by, somebody. Everybody. (x)(Py  (x)((Px & Lxy) & Lyx)) ~(y)((x)(Px  Lyx) v (x)(Px  Lxy)) So now we can do the green “but” statement, which can be treated as an ‘and’ standard form. Go forward when ready for split. Problem 37

Everybody likes, and is liked by, somebody; but nobody likes, or is liked by, everybody. ((x)(Py  (x)((Px & Lxy)&Lyx)) & ~(y)((x)(Px  Lyx) v (x)(Px  Lxy))) Everybody likes, and is liked Nobody likes, or is liked by by, somebody. Everybody. (x)(Py  (x)((Px & Lxy) & Lyx)) ~(y)((x)(Px  Lyx) v (x)(Px  Lxy)) So now we can do the green “but” statement, which can be treated as an ‘and’ standard form. Go forward when ready for split. Problem 37

Nobody gives anything away which is given her by 1 of her children.
New Problem This will take some thinking. Let’s try to think about how it would go as a ‘FET’ statement instead of a ‘NOT:TISS’ statement, although both are possible. Suppose you find someone who has been given something by one of her children. Then what would we say what’s going to happen to that thing? Well, that thing, that same thing, will not be given away. To ‘be given away’ means that there is some person that she gives that thing to. So let’s try to state the ‘FET’ statement. Go forward when ready for restatement. Problem 38

FET:if it is a person then if a child of it gives it something then it will not give that something away. Let’s plug something in for the person, Go forward when ready for plug in. Problem 38

FET:if it is a person then if a child of it gives it something then it will not give that something away. If Ann is a person then if a child of Ann gives Ann something then Ann will not give that something away. Let’s break it up, Go forward when ready for split. Problem 38

FET:if it is a person then if a child of it gives it something then it will not give that something away. If Ann is a person then if a child of Ann gives Ann something then Ann will not give that something away. Ann is a person if a child of Ann gives Ann something then Ann will not give that something away. Ann is a person is easy. The other side still has some complexity to it. Notice that the thing that Ann gets is referred to again, probably on the other side of a horseshoe, so this should be ANYthing Ann gets, a FET statement, and ANY child of Ann. Since there are two FETs here let’s try to do both in next slide. Go forward when ready for restatement. Problem 38

FET:if it is a person then if a child of it gives it something then it will not give that something away. If Ann is a person then if a child of Ann gives Ann something then Ann will not give that something away. Ann is a person if a child of Ann gives Ann something then Pa Ann will not give that something away. FET x:FET y: if x is the child of Ann and x gives Ann y then Ann will not give y away. The red is a restatement of the green with a new notation to save room. Let’s name those two things. Let’s call the child “Rob” and the thing we’ll call “The gift” Go forward when ready for plugin. Problem 38

FET:if it is a person then if a child of it gives it something then it will not give that something away. If Ann is a person then if a child of Ann gives Ann something then Ann will not give that something away. Ann is a person if a child of Ann gives Ann something then Pa Ann will not give that something away. FET x:FET y: if x is the child of Ann and x gives Ann y then Ann will not give y away. If Rob is the child of Ann and Rob gives Ann ‘the gift’ then Ann will not give ‘the gift’ away. But to give the gift away means there is a person that Ann gives the gift to. So with a lot of thought, let’s try symbolizing on the next slide. Let us use Gxyz to mean x gives y to z. Go forward when ready for symbols. Problem 38

FET:if it is a person then if a child of it gives it something then it will not give that something away. If Ann is a person then if a child of Ann gives Ann something then Ann will not give that something away. Ann is a person if a child of Ann gives Ann something then Pa Ann will not give that something away. FET x:FET y: if x is the child of Ann and x gives Ann y then Ann will not give y away. If Rob is the child of Ann and Rob gives Ann ‘the gift’ then Ann will not give ‘the gift’ away. ((Cra & Grga)  ~(x)(Px & Gagx) Now let us take ‘Rob’ and ‘Ann’ out and put in variables for the green statement, which will be the same as the brown statement. Go forward when ready for variables Problem 38

FET:if it is a person then if a child of it gives it something then it will not give that something away. If Ann is a person then if a child of Ann gives Ann something then Ann will not give that something away. Ann is a person if a child of Ann gives Ann something then Pa Ann will not give that something away. (y)(z)((Cya & Gyza)  ~(x)(Px & Gazx) FET x:FET y: if x is the child of Ann and x gives Ann y then Ann will not give y away. If Rob is the child of Ann and Rob gives Ann ‘the gift’ then Ann will not give ‘the gift’ away. ((Cra & Grga)  ~(x)(Px & Gagx) Now we can put together the brown statement Go forward when ready Problem 38

FET:if it is a person then if a child of it gives it something then it will not give that something away. If Ann is a person then if a child of Ann gives Ann something then Ann will not give that something away. (Pa  (y)(z)((Cya & Gyza)  ~(x)(Px & Gazx)) Ann is a person if a child of Ann gives Ann something then Pa Ann will not give that something away. (y)(z)((Cya & Gyza)  ~(x)(Px & Gazx) FET x:FET y: if x is the child of Ann and x gives Ann y then Ann will not give y away. If Rob is the child of Ann and Rob gives Ann ‘the gift’ then Ann will not give ‘the gift’ away. ((Cra & Grga)  ~(x)(Px & Gagx) Now Ann goes out and we need a new variable. How about ‘x1’ to represent the green statement and then since the brown is synonymous, it would go the same way Go forward when ready for final step. Problem 38

(x1)(Px1  (y)(z)((Cyx1 & Gyzx1)  ~(x)(Px & Gx1zx)) FET:if it is a person then if a child of it gives it something then it will not give that something away. If Ann is a person then if a child of Ann gives Ann something then Ann will not give that something away. (Pa  (y)(z)((Cya & Gyza)  ~(x)(Px & Gazx)) Ann is a person if a child of Ann gives Ann something then Pa Ann will not give that something away. (y)(z)((Cya & Gyza)  ~(x)(Px & Gazx) FET x:FET y: if x is the child of Ann and x gives Ann y then Ann will not give y away. If Rob is the child of Ann and Rob gives Ann ‘the gift’ then Ann will not give ‘the gift’ away. ((Cra & Grga)  ~(x)(Px & Gagx) A very hard problem. There were other ways to approach this. We could have focused upon the word ‘nobody’ and realized that this means ‘it is not the case that there exists someone such that’. Go forward to see an alternative. Problem 38

(x1)(Px1  (y)(z)((Cyx1 & Gyzx1)  ~(x)(Px & Gx1zx)) FET:if it is a person then if a child of it gives it something then it will not give that something away. alternately NOT:TISS:it is a person it is given something from one of its children and it gives that thing away to someone. ~(Ex)(Px&(Ey)(Cyx & (Ez)(Gyzx & (Ex1)(Px1 &Gxzx1))))) These statements are logically equivalent and both correct. As an aside “To give something AWAY” may imply that you are not giving it to yourself and may include some assumption about not giving it back to the person who gave it to you. If so, then a better answer would involve identity. Go forward for a new problem. Problem 38

Only Philosophers and Logicians understand everything. New Problem
Again we have an ‘only’ deal with and we realize that it is not the same as ‘All’, The ‘All’ statement would be FET: if it is a philosopher or a logician then it understands everything. Let’s avoid that though, and realize that some switching is likely to occur. Imagine you found something that understood everything, then what could you say about it? Go forward to see statement in standard form. Problem 39

Only Philosophers and Logicians understand everything.
FET:if it understands everything then it is a philosopher or a logician. This is different from the ‘all’ statement. To better understand this let us plug some example thing in, like ‘Art’. Go forward to see plug in. Problem 39

FET:if it understands everything then it is a philosopher or a logician. If Art understands everything then Art is a philosopher or a logician. Let’s break it up. Think about it and go forward for breakup. Problem 39

FET:if it understands everything then it is a philosopher or a logician. If Art understands everything then Art is a philosopher or a logician. Art understands everything Art is a philosopher or a logician. The right side is easy and the left side is not so bad. Let’s symbolize the right side and put the left side in standard form. Think about it and go forward when ready. Problem 39

FET:if it understands everything then it is a philosopher or a logician. If Art understands everything then Art is a philosopher or a logician. Art understands everything Art is a philosopher or a logician. (Pa v La) FET: Art understands it. Usually a FET has an “if..then” underneath. This one is even simpler Think about it and go forward when ready. Problem 39

FET:if it understands everything then it is a philosopher or a logician. If Art understands everything then Art is a philosopher or a logician. Art understands everything Art is a philosopher or a logician. (Pa v La) FET: Art understands it. (x)Uax We should be able to do the green and the brown in one slide. The green is synonymous with the FET statement and the brown is just a ‘if..then’ statement. Think about it and go forward when ready. Problem 39

FET:if it understands everything then it is a philosopher or a logician. If Art understands everything then Art is a philosopher or a logician. ((x)Uax  (Pa v La)) Art understands everything Art is a philosopher or a logician. (x)Uax (Pa v La) FET: Art understands it. (x)Uax We should be able to do the green and the brown in one slide. The green is a FET statement and the brown is just synonymous with that. Think about it and go forward when ready. Problem 39

(y)((x)Uyx  (Py v Ly)) FET:if it understands everything then it is a philosopher or a logician. If Art understands everything then Art is a philosopher or a logician. ((x)Uax  (Pa v La)) Art understands everything Art is a philosopher or a logician. (x)Uax (Pa v La) FET: Art understands it. (x)Uax Go forward when you are ready for a new problem. Problem 39

Only Philosophers and Logicians understand anything.
New Problem Very much like the last problem. The problem before talked about ‘everything’, this one talks about ‘anything’ What is the difference? Go forward when you are ready to put it in standard form. Problem 40

FET:if it understands anything then it is a philosopher or a Logician. Let’s pick an instance. Go forward. Problem 40

FET:if it understands anything then it is a philosopher or a Logician. If Al understands anything then Al is a philosopher or a Logician. Is this ‘if..then’ statement in standard form? Not really because Al understands anything is not an English statement But there are a couple of ways to put this in standard form. Think about it and go forward. Problem 40

FET:if it understands anything then it is a philosopher or a logician. If Al understands anything then Al is a philosopher or a logician. If Al understands something then Al is a philosopher or a logician. Now it is in standard form and we should be able to turn it into symbols. Last time we had if Art understands everything then Art is a philosopher or a logician. This is just a bit different. Try symbolizing and then go forward. Problem 40

FET:if it understands anything then it is a philosopher or a logician. If Al understands anything then Al is a philosopher or a logician. If Al understands something then Al is a philosopher or a logician. ((x)Uax  (Pa v La)) Notice that this is not an existential over a horseshoe, but a horseshoe statement with an existential the left side. The green statement says the same thing, and so it will be represented the same way. Go forward when ready. Problem 40

FET:if it understands anything then it is a philosopher or a logician. If Al understands anything then Al is a philosopher or a logician. ((x)Uax  (Pa v La)) If Al understands something then Al is a philosopher or a logician. For the green FET statement we picked Al, so let us use y to replace Al for the green. And then of course the brown is synonymous with the green. Go forward when ready. Problem 40

(y)((x)Uyx  (Py v Ly)) FET:if it understands anything then it is a philosopher or a logician. If Al understands anything then Al is a philosopher or a logician. ((x)Uax  (Pa v La)) If Al understands something then Al is a philosopher or a logician. There was another logically equivalent way to do this. For the brown statement we could have said “for each thing, if Al understands it, then Al is a philosopher or a logician” . Let us see how that would look. Go forward when ready. Problem 40

(y)((x)Uyx  (Py v Ly)) (x)(y)(Uxy  (Px v Lx)) FET:if it understands something then it is a philosopher or a logician. FET(x) FET(y): If x understands y then x is either a philosopher or a logician. Which one is more intuitive to you? Go forward when ready for new problem. Problem 40

Whenever a lecher sees a girl, he gets excited.
New Problem The thing that makes this problem somewhat interesting is that in introduces time. ‘Whenever’ means ‘any time’. So instead of a dictionary like S_.:_sees. We need something like ‘x sees y at time z’ So let’s restate this with any lecher and any time on the next slide. Go forward when ready. Problem 41

FET(x):FET(y):if x is a lecher and y is a time and x sees a girl at that time then x gets excited at that time. Let’s plug in a couple names. Chester and 1.pm Go forward when ready. Problem 41

FET(x):FET(y):if x is a lecher and y is a time and x sees a girl at time y then x gets excited at time y. If Ches is a lecher and 1 pm is a time and Ches sees a girl at 1 then Ches gets excited at 1. Let’s split this ‘if..then’ statement. Go forward when ready. Problem 41

FET(x):FET(y):if x is a lecher and y is a time and x sees a girl at time y then x gets excited at time y. If Ches is a lecher and 1 pm is a time and Ches sees a girl at 1 then Ches gets excited at 1. Ches is a lecher and 1 pm is a time and Ches sees a girl at Ches gets excited at 1. Let us take ‘sees a girl’ to mean there is at least one girl he sees at one pm.Can we turn these into symbols now? I hope so. Go forward when ready. Problem 41

FET(x):FET(y):if x is a lecher and y is a time and x sees a girl at time y then x gets excited at time y. If Ches is a lecher and 1 pm is a time and Ches sees a girl at 1 then Ches gets excited at 1. Ches is a lecher and 1 pm is a time and Ches sees a girl at Ches gets excited at 1. (Lc & (To & (z)(Gz & Sczo))) Eco We have enough experience now, let’s do all three steps on next slide. Build the ‘if..then’ statement (green), then build the double universal statement, (brown), which is equivalent to the original (black) Go forward when ready. Problem 41

(x)(y)((Lx& (Ty & (z)(Gz & Sxzy)))  Exy) FET(x):FET(y):if x is a lecher and y is a time and x sees a girl at time y then x gets excited at time y. If Ches is a lecher and 1 pm is a time and Ches sees a girl at 1 then Ches gets excited at 1. ((Lc & (To & (z)(Gz & Sczo)))  Eco) Ches is a lecher and 1 pm is a time and Ches sees a girl at Ches gets excited at 1. (Lc & (To & (z)(Gz & Sczo))) Eco We have enough experience now, let’s do all three steps on next slide. Build the ‘if..then’ statement (green), then build the double universal statement, (brown), which is equivalent to the original (black) Go forward when ready for a new problem. Problem 41

Whither thou goest, I will go.
New Problem What this seems to say in an older English is ‘wherever you go, I will go. So this is ‘for any PLACE’. This one is not too hard once you get past the English Go forward when ready for standard form. Problem 42

Whither thou goest, I will go.
FET:if it is a place and you go to it then I go to it. Let’s finish off in one slide. Go forward when ready for standard form. Problem 42

Whither thou goest, I will go. (x)((Px & Gux)  Gix)
FET:if it is a place and you go to it then I go to it. Hard to see why this is a later problem. It doesn’t seem that hard. Go forward when ready for new problem. Problem 42

No one on A’s team can beat everyone on C’s team. New Problem
This gets interesting. It is probably easiest to consider this a NOT:TISS statement How would we do that? Go forward when ready for NOT:TISS form. Problem 43

No one on A’s team can beat everyone on C’s team.
NOT:TISS:it is a person, it is on A’s team and it can beat everyone on C’s team. Let’s rip off the ‘NOT’ and plug in ‘Al’ for the person. Go forward when ready for rip and plug in. Problem 43

NOT:TISS:it is a person, it is on A’s team and it can beat everyone on C’s team. Al is a person, Al is on A’s team and Al can beat everyone on C’s team. Let’s consider ‘and’ as the standard form connector and break it there. Go forward when ready for break. Problem 43

NOT:TISS:it is a person, it is on A’s team and it can beat everyone on C’s team. Al is a person, Al is on A’s team and Al can beat everyone on C’s team. Al is a person, Al is on A’s team Al can beat everyone on C’s team. So we just have to do the left, which is an ampersand statement and figure how to do the right. Go forward when ready for symbols. Problem 43

NOT:TISS:it is a person, it is on A’s team and it can beat everyone on C’s team. Al is a person, Al is on A’s team and Al can beat everyone on C’s team. Al is a person, Al is on A’s team Al can beat everyone on C’s team. (Pa & Aa) (x)((Px & Cx)  Bax) Let’s build the green statement. Go forward when ready for symbols. Problem 43

NOT:TISS:it is a person, it is on A’s team and it can beat everyone on C’s team. Al is a person, Al is on A’s team and Al can beat everyone on C’s team. ((Pa & Aa) & (x)((Px & Cx)  Bax)) Al is a person, Al is on A’s team Al can beat everyone on C’s team. (Pa & Aa) (x)((Px & Cx)  Bax) Al was just the ‘something’ so let’s put in a new variable and build the green ‘not’ statement in one step. And of course the green and the brown are equivalent, so they will look the same, Go forward when ready for green building. Problem 43

(y)((Py & Ay) & (x)((Px & Cx)  Byx)) NOT:TISS:it is a person, it is on A’s team and it can beat everyone on C’s team. Al is a person, Al is on A’s team and Al can beat everyone on C’s team. ((Pa & Aa) & (x)((Px & Cx)  Bax)) Al is a person, Al is on A’s team Al can beat everyone on C’s team. (Pa & Aa) (x)((Px & Cx)  Bax) There is another way to do it. You could say that everyone on A’s team has someone on C’s team that they cannot beat. See if you can do that in one step. Go forward when ready for alternative. Problem 43

(y)((Py & Ay) & (x)((Px & Cx)  Byx)) (x)((Py & Ay )  (x)((Px & Cx) & ~Byx)) NOT:TISS:it is a person, it is on A’s team and it can beat everyone on C’s team. FET:if it is a person on A’s team then there is a person on C’s team he can’t beat. These are logically equivalent and thus both correct if one of them is. (And at least one of them is.) Go forward when ready for new problem. Problem 43

No one on A’s team can beat anyone on C’s team. New Problem
This is similar to the last one where only one word changes. ‘Everyone’ gets changed to ‘Anyone’ Let’s see what happens when if we treat this like the other, as a ‘NOT:TISS:’ statement. Go forward when ready for standard form. Problem 44

No one on A’s team can beat anyone on C’s team.
NOT:TISS:It is a person on A’s team and it can beat anyone on C’s team. But this is wrong because were denying that there is a person who can beat anyone on C’s team, but instead that he can beat someone on C’s team. Let us restate it better. Go forward when ready for replacement. Problem 44

NOT:TISS:It is a person on A’s team and it can beat someone on C’s team. Go forward when ready for ‘NOT rip’ and plug in. Problem 44

NOT:TISS:It is a person on A’s team and it can beat someone on C’s team. Al is a person on A’s team and Al can beat someone on C’s team. Let’s do the symbols now on this one. Go forward when ready for symbolization. Problem 44

NOT:TISS:It is a person on A’s team and it can beat someone on C’s team. Al is a person on A’s team and Al can beat someone on C’s team. ((Pa & Aa) & (x)((Px & Cx) & Bax)) We should be able to do the green and brown now. Go forward when ready for final answer. Problem 44

~(y)((Py & Ay) & (x)((Px & Cx) & Byx)) NOT:TISS:It is a person on A’s team and it can beat someone on C’s team. Al is a person on A’s team and Al can beat someone on C’s team. ((Pa & Aa) & (x)((Px & Cx) & Bax)) There is an alternative to this representation, just as there was with the last one. Note that if you had a person on A’s team and one on C’s team then the one on A’s team could not beat the one on C’s team, for any pick. So let’s just to that symbolization Go forward when ready for alternate. Problem 44

~(y)((Py & Ay) & (x)((Px & Cx) & Byx)) (y)((Py & Ay)  ( x)((Px & Cx)  ~Byx)) NOT:TISS:It is a person on A’s team and it can beat someone on C’s team. FET:if it is a person on A’s team then it cannot beat anyone on C’s team. (y)((Py & Ay)  (x)((Px & Cx)  ~Byx)) Alternately (y)(x)(((Py & Ay) & (Px & Cx)) ~Byx)) The last is logically equivalent, but it would have to come from a slightly different standard form. Go forward when ready for new problem. Problem 44

Go forward when ready to view standard form.
Anyone on A’s team that can beat someone on C’s team can beat everyone on D’s team. New Problem This seems straightforwardly a FET statement. Let’s just change the minimum and express this as a FET standard form. Go forward when ready to view standard form. Problem 45

Anyone on A’s team that can beat someone on C’s team can beat everyone on D’s team. FET:if it is a person on A’s team and it can beat someone on C’s team then it can beat everyone on D’s team Let’s plug in Al Go forward when ready to view standard form. Problem 45

Anyone on A’s team that can beat someone on C’s team can beat everyone on D’s team. FET:if it is a person on A’s team and it can beat someone on C’s team then it can beat everyone on D’s team. If Al is a person on A’s team and Al can beat someone on C’s team then Al can beat everyone on D’s team. Let’s break it up. Go forward when ready to view standard form. Problem 45

Al is a person on A’s team and Al can beat everyone on D’s team.
Anyone on A’s team that can beat someone on C’s team can beat everyone on D’s team. FET:if it is a person on A’s team and it can beat someone on C’s team then it can beat everyone on D’s team. If Al is a person on A’s team and Al can beat someone on C’s team then Al can beat everyone on D’s team. Al is a person on A’s team and Al can beat everyone on D’s team. Al can beat someone on C’s team Let’s symbolize each side. Go forward when ready to view standard form. Problem 45

(y)(((Py & Ay) & (x)((Px & Cx)& Byx)) (x)((Px& Dx)  Byx))
Anyone on A’s team that can beat someone on C’s team can beat everyone on D’s team. (y)(((Py & Ay) & (x)((Px & Cx)& Byx)) (x)((Px& Dx)  Byx)) FET:if it is a person on A’s team and it can beat someone on C’s team then it can beat everyone on D’s team. If Al is a person on A’s team and Al can beat someone on C’s team then Al can beat everyone on D’s team. (((Pa & Aa) & (x)((Px & Cx)& Bax)) (x)((Px& Dx)  Bax)) Al is a person on A’s team and Al can beat everyone on D’s team. Al can beat someone on C’s team (x)((Px& Dx)  Bax) ((Pa & Aa) & (x)((Px & Cx)& Bax)) A fairly complex problem but manageable once we broke it up a bit. Go forward when ready for new problem. Problem 45

Peter once had some wits but Mack never had any. New Problem
The complexity of this one has to do with the dictionary. Let it be that Hxyz: means x has y at time z, Wx:x is a wit, Tx: x is a time and of course p:peter, m:mac. It’s a ‘but’ statement that is synonymous with an ‘and’ standard form. So let’s restate it in standard form. Go forward when ready for breakup. Problem 46

Peter once had some wits but Mack never had any.
Peter once had some wits and Mack never had any wits. Let’s break it up Go forward when ready for breakup. Problem 46

Peter once had some wits and Mack never had any wits. Peter once had some wits Mack never had any wits. What do these mean? Let’s try to restate each red in standard form. Go forward when ready for standard form. Problem 46

Peter once had some wits and Mack never had any wits. Peter once had some wits Mack never had any wits. TISS:it is a time and NOT:TISS:it is a time and Peter had a wit at that time Mack had a wit at that time. On the right side we could have said for all times and for all wits, it is not the case that Mack had that wit at that time’ but this will do also. Lets turn them into symbols and finish it off in one slide. Go forward when ready for symbolization. Problem 46

((x)(Tx & (y)(Wy & Hpyx)) & ~(x)(Tx & (y)(Wy & Hmyx))) Peter once had some wits and Mack never had any wits. Peter once had some wits Mack never had any wits. (x)(Tx & (y)(Wy & Hpyx)) ~(x)(Tx & (y)(Wy & Hmyx)) TISS:it is a time and NOT:TISS:it is a time and Peter had a wit at that time Mack had a wit at that time. Alternately (x)(Tx  (y)(Wy  ~Hmyx)) . Go forward when ready for new problem. Problem 46

A thing is a part of another thing iff whatever overlaps the first also overlaps the second.
New Problem Let us to this as a double FET statement. Go forward when ready for double FET. Problem 47

FET x:FET y: x is part of y iff whatever overlaps x overlaps y.
A thing is a part of another thing iff whatever overlaps the first also overlaps the second. FET x:FET y: x is part of y iff whatever overlaps x overlaps y. Now let us pick some example things, like Ohio and the U.S. Go forward when ready for instances. Problem 47

A thing is a part of another thing iff whatever overlaps the first also overlaps the second. FET x:FET y: x is part of y iff whatever overlaps x overlaps y. Ohio is part of the U.S. iff whatever overlaps Ohio overlaps the U.S. Let’s break it into its parts. Go forward when ready for breakdown. Problem 47

A thing is a part of another thing iff whatever overlaps the first also overlaps the second. FET x:FET y: x is part of y iff whatever overlaps x overlaps y. Ohio is part of the U.S. iff whatever overlaps Ohio overlaps the U.S. Ohio is part of the U.S Whatever overlaps Ohio overlaps the U.S. The left side is easy. Let’s symbolize it and state the right as an FET statement Go forward when ready. Problem 47

A thing is a part of another thing iff whatever overlaps the first also overlaps the second. FET x:FET y: x is part of y iff whatever overlaps x overlaps y. Ohio is part of the U.S. iff whatever overlaps Ohio overlaps the U.S. Ohio is part of the U.S Whatever overlaps Ohio overlaps the U.S. Pou FET:if it overlaps Ohio then it overlaps the U.S. Now let’s symbolize the right red statement Go forward when ready. Problem 47

A thing is a part of another thing iff whatever overlaps the first also overlaps the second. FET x:FET y: x is part of y iff whatever overlaps x overlaps y. Ohio is part of the U.S. iff whatever overlaps Ohio overlaps the U.S. Ohio is part of the U.S Whatever overlaps Ohio overlaps the U.S. Pou FET:if it overlaps Ohio then it overlaps the U.S. (z)(Ozo  Ozu) The red ‘FET’ statement is equivalent to the green so the green should be represented the same way. Go forward when ready. Problem 47

A thing is a part of another thing iff whatever overlaps the first also overlaps the second. FET x:FET y: x is part of y iff whatever overlaps x overlaps y. Ohio is part of the U.S. iff whatever overlaps Ohio overlaps the U.S. Ohio is part of the U.S Whatever overlaps Ohio overlaps the U.S. Pou (z)(Ozo  Ozu) FET:if it overlaps Ohio then it overlaps the U.S. (z)(Ozo  Ozu) The green is an ‘iff’ statement and a double instance of the brown which is synonymous with the black. So let’s finish up the representation in one slide. Go forward when ready for final answer. Problem 47

(x)(y)(Pxy  (z)(Ozx  Ozy))
A thing is a part of another thing iff whatever overlaps the first also overlaps the second. (x)(y)(Pxy  (z)(Ozx  Ozy)) FET x:FET y: x is part of y iff whatever overlaps x overlaps y. Ohio is part of the U.S. iff whatever overlaps Ohio overlaps the U.S. (Pou  (z)(Ozo  Ozu)) Ohio is part of the U.S Whatever overlaps Ohio overlaps the U.S. Pou (z)(Ozo  Ozu) FET:if it overlaps Ohio then it overlaps the U.S. (z)(Ozo  Ozu) An interesting axiom connecting ‘part’ to ‘overlap’. It could even be a definition. Go forward when ready for new problem. Problem 47

Go forward when ready double FET.
If a thing overlaps all the parts of another thing, then the second is part of the first. New Problem Again we have a double FET statement. Let’s restate it in a double standard form. Go forward when ready double FET. Problem 48

FET x: FET y: If x overlaps all the parts of y, then y is part of x.
If a thing overlaps all the parts of another thing, then the second is part of the first. FET x: FET y: If x overlaps all the parts of y, then y is part of x. Let’s pick California and L.A. as our example things Go forward when ready for new problem. Problem 48

If a thing overlaps all the parts of another thing, then the second is part of the first. FET x: FET y: If x overlaps all the parts of y, then y is part of x. If California overlaps all the parts of L.A., then L.A is part of California An ‘if..then’ form. Let’s break it up Go forward when ready. Problem 48

If a thing overlaps all the parts of another thing, then the second is part of the first. FET x: FET y: If x overlaps all the parts of y, then y is part of x. If California overlaps all the parts of L.A., then L.A is part of California California overlaps all the parts of L.A L.A is part of California The right red is easy. The left red could be stated as a ‘FET’ statement. Let’s symbolize the right and change the left into standard form. Go forward when ready. Problem 48

If a thing overlaps all the parts of another thing, then the second is part of the first. FET x: FET y: If x overlaps all the parts of y, then y is part of x. If California overlaps all the parts of L.A., then L.A is part of California California overlaps all the parts of L.A L.A is part of California Pcl FET:if it is a part of L.A. the California overlaps it Now let’s symbolize the left. Go forward when ready. Problem 48

If a thing overlaps all the parts of another thing, then the second is part of the first. FET x: FET y: If x overlaps all the parts of y, then y is part of x. If California overlaps all the parts of L.A., then L.A is part of California California overlaps all the parts of L.A L.A is part of California Pcl FET:if it is a part of L.A. the California overlaps it (z)(Pzl  Ocz) This red FET statement is synonymous with the green, so now let’s do the green. Go forward when ready. Problem 48

If a thing overlaps all the parts of another thing, then the second is part of the first. FET x: FET y: If x overlaps all the parts of y, then y is part of x. If California overlaps all the parts of L.A., then L.A is part of California California a overlaps all the parts of L.A L.A is part of California (z)(Pzl  Ocz) Plc FET:if it is a part of L.A. the California overlaps it (z)(Pzl  Ocz) So now the reds come from the ‘if..then’ green which comes from the double FET brown which comes from the synonymous top black. So let’s do it all in one slide. Go forward when ready. Problem 48

(x)(y)((z)(Pzy  Oxz)  Pyx)
If a thing overlaps all the parts of another thing, then the second is part of the first. (x)(y)((z)(Pzy  Oxz)  Pyx) FET x: FET y: If x overlaps all the parts of y, then y is part of x. If California overlaps all the parts of L.A., then L.A is part of California ((z)(Pzl  Ocz)  Plc) California a overlaps all the parts of L.A L.A is part of California (z)(Pzl  Ocz) Plc FET:if it is a part of L.A. the California overlaps it (z)(Pzl  Ocz) At this point we may be able to appreciate how the process helps us get the answer. It might have been hard to anticipate the universal z on the left side of a horseshoe without the process. Go forward when ready for new problem. Problem 48

A double ‘FET’ let’s state it in double standard form.
A thing overlaps another thing, iff, the two things have a part in common. New Problem A double ‘FET’ let’s state it in double standard form. Go forward when ready. Problem 49

FET x:FET y: x overlaps y iff x and y have a part in common.
A thing overlaps another thing, iff, the two things have a part in common. FET x:FET y: x overlaps y iff x and y have a part in common. Let’s plug in a couple of things. How about the senate membership and the house membership. Go forward when ready. Problem 49

A thing overlaps another thing, iff, the two things have a part in common. FET x:FET y: x overlaps y iff x and y have a part in common. The Senate overlaps The House iff The Senate and The House have a part in common. It’s in standard form, let’s break it up. Go forward when ready. Problem 49

A thing overlaps another thing, iff, the two things have a part in common. FET x:FET y: x overlaps y iff x and y have a part in common. The Senate overlaps The House iff The Senate and The House have a part in common. The Senate overlaps The House The Senate and The House have a part in common. The left red is easy. What does the right red mean? What does it mean to have a part in common? Go forward when ready to put right red in standard form. Problem 49

A thing overlaps another thing, iff, the two things have a part in common. FET x:FET y: x overlaps y iff x and y have a part in common. The Senate overlaps The House iff The Senate and The House have a part in common. The Senate overlaps The House The Senate and The House have a part in common. Osh TISS: it is part of the Senate and it is part of the House To have a part in common means there is something such that it is part of the first thing and it is part of the second thing. Now we can turn the red TISS statement into symbols. Go forward when ready. Problem 49

A thing overlaps another thing, iff, the two things have a part in common. FET x:FET y: x overlaps y iff x and y have a part in common. The Senate overlaps The House iff The Senate and The House have a part in common. The Senate overlaps The House The Senate and The House have a part in common. Osh TISS: it is part of the Senate and it is part of the House. (z)(Pzs & Pzh) This ‘TISS’ red statement is synonymous with the green. So let’s do the green. Go forward when ready. Problem 49

A thing overlaps another thing, iff, the two things have a part in common. FET x:FET y: x overlaps y iff x and y have a part in common. The Senate overlaps The House iff The Senate and The House have a part in common. The Senate overlaps The House The Senate and The House have a part in common. Osh (z)(Pzs & Pzh) TISS: it is part of the Senate and it is part of the House. (z)(Pzs & Pzh) The reds build the ‘iff’ green which builds the double ‘FET’ brown which builds the synonymous black statement. Go forward when ready for final answer. Problem 49

A thing overlaps another thing, iff, the two things have a part in common. (x)(y)(Oxy  (z)(Pzx & Pzy)) FET x:FET y: x overlaps y iff x and y have a part in common. The Senate overlaps The House iff The Senate and The House have a part in common. (Osh  (z)(Pzs & Pzh)) The Senate overlaps The House The Senate and The House have a part in common. Osh (z)(Pzs & Pzh) TISS: it is part of the Senate and it is part of the House. (z)(Pzs & Pzh) So essentially it says two things overlap iff there is something such that it is part of the first and it is part of the second. Go forward when ready for new problem. Problem 49

Go forward when ready for 50a.
You, (I.e. any person) can fool some of the people all of the time, and all of the people some of the time, but you cannot fool all of the people all of the time. New Problem There are several ways to do this and there is some ambiguity, but the easiest in the room we have is to consider it as having three parts connected by ‘ands’ (or ‘buts’ which make synonymous forms). The term ‘You’ meaning ‘any person’ stretches all the way to the ‘but’, however a ‘FET’ statement over an ‘and’ is synonymous to an ‘and’ statement with ‘FET’ statements on each side. So let us do this as problem 50a, 50b and 50c and put them together later on. Go forward when ready for 50a. Problem 50

You can fool some of the people all of the time.
It looks like we have to make a decision here. Does this say that there is one guy out there and you can fool him at any time? Or does it say any time you pick you can find someone to fool at that time? Maybe we don’t have to make a decision now. Let us just restate this as a ‘FET’ statement and see where it goes. Go forward when ready for ‘FET’ standard form. Problem 50a

FET:if it is a person then it can fool some of the people all of the time. Let’s pick ‘Bush’ Go forward when ready for instance. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Let’s break it up Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Bush is a person Bush can fool some of the people all of the time. The left is easy. But did the ambiguity disappear on the right? It seems to have. It seems clear that this means that there is at least one person Bush can fool all the time. Let’s represent the left and put the right in standard form. Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Bush is a person Bush can fool some of the people all of the time. Pb TISS:It is a person and Bush can fool it all of the time. Let’s plug in a person. Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Bush is a person Bush can fool some of the people all of the time. Pb TISS:It is a person and Bush can fool it all of the time. Laura is a person and Bush can fool Laura all of the time. Let’s break it up Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Bush is a person Bush can fool some of the people all of the time. Pb TISS:It is a person and Bush can fool it all of the time. Laura is a person and Bush can fool Laura all of the time. Laura is a person Bush can fool Laura all of the time. The right talks about ‘all times’ it’s going to be another quantifier. But now we should be able to turn this into symbols. Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Bush is a person Bush can fool some of the people all of the time. Pb TISS:It is a person and Bush can fool it all of the time. Laura is a person and Bush can fool Laura all of the time. Laura is a person Bush can fool Laura all of the time. Pl (z)(Tz  Fblz) Let’s put the green together Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Bush is a person Bush can fool some of the people all of the time. Pb TISS:It is a person and Bush can fool it all of the time. Laura is a person and Bush can fool Laura all of the time. (Pl & (z)(Tz  Fblz)) Laura is a person Bush can fool Laura all of the time. Pl (z)(Tz  Fblz) Laura was just an instance of the green. Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Bush is a person Bush can fool some of the people all of the time. Pb TISS:It is a person and Bush can fool it all of the time. (y)(Py & (z)(Tz  Fbyz)) Laura is a person and Bush can fool Laura all of the time. (Pl & (z)(Tz  Fblz)) Laura is a person Bush can fool Laura all of the time. Pl (z)(Tz  Fblz) The red is synonymous with the green. So let’s do the green. Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. Bush is a person Bush can fool some of the people all of the time. Pb (y)(Py & (z)(Tz  Fbyz)) TISS:It is a person and Bush can fool it all of the time. (y)(Py & (z)(Tz  Fbyz)) Laura is a person and Bush can fool Laura all of the time. (Pl & (z)(Tz  Fblz)) Laura is a person Bush can fool Laura all of the time. Pl (z)(Tz  Fblz) Let’s put the green ‘if..then’ statement together. Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. (Pb (y)(Py & (z)(Tz  Fbyz))) Bush is a person Bush can fool some of the people all of the time. Pb (y)(Py & (z)(Tz  Fbyz)) TISS:It is a person and Bush can fool it all of the time. (y)(Py & (z)(Tz  Fbyz)) Laura is a person and Bush can fool Laura all of the time. (Pl & (z)(Tz  Fblz)) Laura is a person Bush can fool Laura all of the time. Pl (z)(Tz  Fblz) Bush was just an instance of the green, so let’s put in the variable x. Go forward when ready. Problem 50a

FET: If it is a person then it can fool some of the people all of the time. (x)(Px (y)(Py & (z)(Tz  Fxyz))) If Bush is a person then Bush can fool some of the people all of the time. (Pb (y)(Py & (z)(Tz  Fbyz))) Bush is a person Bush can fool some of the people all of the time. Pb (y)(Py & (z)(Tz  Fbyz)) TISS:It is a person and Bush can fool it all of the time. (y)(Py & (z)(Tz  Fbyz)) Laura is a person and Bush can fool Laura all of the time. (Pl & (z)(Tz  Fblz)) Laura is a person Bush can fool Laura all of the time. Pl (z)(Tz  Fblz) It is synonymous with the green so it gets represented the same way. Go forward when ready. Problem 50a

(x)(Px (y)(Py & (z)(Tz  Fxyz))) FET: If it is a person then it can fool some of the people all of the time. If Bush is a person then Bush can fool some of the people all of the time. (Pb (y)(Py & (z)(Tz  Fbyz))) Bush is a person Bush can fool some of the people all of the time. Pb (y)(Py & (z)(Tz  Fbyz)) TISS:It is a person and Bush can fool it all of the time. (y)(Py & (z)(Tz  Fbyz)) Laura is a person and Bush can fool Laura all of the time. (Pl & (z)(Tz  Fblz)) Laura is a person Bush can fool Laura all of the time. Pl (z)(Tz  Fblz) There you have one third of the problem. Go forward when ready for 50b. Problem 50a

You can fool all of the people some of the time.
There is an seeming ambiguity in this statement like 50a. Does it mean that there is some time and you can fool everyone at that time? Or does it mean that every person has a time you can fool him, possibly a different time for each. Let us see if this seeming ambiguity disappears upon analysis like it did with 50a. Go forward when ready ‘FET’ statement. Problem 50b

FET:If it is a person then it can fool all of the people some of the time. Again let’s pick on Bush. Go forward when ready. Problem 50b

FET:If it is a person then it can fool all of the people some of the time. If Bush is a person then Bush can fool all of the people some of the time. Let’s break it up. Go forward when ready. Problem 50b

FET:If it is a person then it can fool all of the people some of the time. If Bush is a person then Bush can fool all of the people some of the time. Bush is a person Bush can fool all of the people some of the time. The left side is atomic and easy. Have we cleared up the problem with the right side. Bush can fool all of the people some of the time.It seems to me that it says there is a time when Bush can fool all those people at that time. I think if you wanted to say that he fools them at possibly different times you would say ‘Bush can fool everybody sometime.’ So let’s represent the left and make the right a ‘TISS’ statement Go forward when ready. Problem 50b

FET:If it is a person then it can fool all of the people some of the time. If Bush is a person then Bush can fool all of the people some of the time. Bush is a person Bush can fool all of the people some of the time. Pb TISS:it is a time and Bush can fool all of the people at it (that time). With our experience now, we should be able to turn this red ‘TISS’ statement into a formal statement. Go forward when ready to symbolize. Problem 50b

FET:If it is a person then it can fool all of the people some of the time. If Bush is a person then Bush can fool all of the people some of the time. Bush is a person Bush can fool all of the people some of the time. Pb TISS:it is a time and Bush can fool all of the people at it (that time). (y)(Ty & (z)(Pz  Fbzy)) This ‘TISS’ statement is synonymous with the green, so lets do the green. Go forward when ready. Problem 50b

FET:If it is a person then it can fool all of the people some of the time. If Bush is a person then Bush can fool all of the people some of the time. Bush is a person Bush can fool all of the people some of the time. Pb (y)(Ty & (z)(Pz  Fbzy)) TISS:it is a time and Bush can fool all of the people at it (that time). (y)(Ty & (z)(Pz  Fbzy)) Let’s build the ‘if..then’ green statement. Go forward when ready. Problem 50b

FET:If it is a person then it can fool all of the people some of the time. If Bush is a person then Bush can fool all of the people some of the time. (Pb (y)(Ty & (z)(Pz  Fbzy)) Bush is a person Bush can fool all of the people some of the time. Pb (y)(Ty & (z)(Pz  Fbzy)) TISS:it is a time and Bush can fool all of the people at it (that time). (y)(Ty & (z)(Pz  Fbzy)) Let’s build the ‘FET’ green statement and then the brown, which is synonymous with the green. Go forward when ready. Problem 50b

(x)(Px (y)(Ty & (z)(Pz  Fxzy)) FET:If it is a person then it can fool all of the people some of the time. If Bush is a person then Bush can fool all of the people some of the time. (Pb (y)(Ty & (z)(Pz  Fbzy)) Bush is a person Bush can fool all of the people some of the time. Pb (y)(Ty & (z)(Pz  Fbzy)) TISS:it is a time and Bush can fool all of the people at it (that time). (y)(Ty & (z)(Pz  Fbzy)) Now we’ve done two parts to problem 50. Let’s see if we can do the third. Go forward when ready for 50c. Problem 50b

You cannot fool all of the people all of the time.
There may be an ambiguity here but it’s not compelling. Could this mean that ‘Not everyone can fool everyone all the time’? Or does it mean ‘No one can fool everyone all the time’? It seems to be the latter. This should be simple. Let us try it as a ‘NOT:TISS:’ Go forward when ready standard form. Problem 50c

NOT:TISS:it is a person and it can fool everyone all the time. Let’s remove the ‘NOT’ and plug in ‘Bush’. Go forward when ready standard form. Problem 50c

NOT:TISS:it is a person and it can fool everyone all the time. Bush is a person and Bush can fool everyone all the time. Let’s break it up.. Go forward when ready standard form. Problem 50c

NOT:TISS:it is a person and it can fool everyone all the time. Bush is a person and Bush can fool everyone all the time. Bush is a person Bush can fool everyone all the time. The left is atomic and easy. The right is a double quantified statement. Let us symbolize the left and restate the right in standard form. Go forward when ready standard form. Problem 50c

NOT:TISS:it is a person and it can fool everyone all the time. Bush is a person and Bush can fool everyone all the time. Bush is a person Bush can fool everyone all the time. Pb FET y:FET z: If y is a person and z is a time then Bush can fool y at z It could have been done a few other ways, one of which was to say ‘FET’ if it is a person then Bush can fool it all the time. This would also work. I use the double FET to same space. Let’s symbolize this double ‘FET’ statement on the next slide. Go forward when ready standard form. Problem 50c

NOT:TISS:it is a person and it can fool everyone all the time. Bush is a person and Bush can fool everyone all the time. Bush is a person Bush can fool everyone all the time. Pb FET y:FET z: If y is a person and z is a time then Bush can fool y at z (y)(z)((Py & Tz)  Fbyz) This just means the same as the green, so let’s do the green Go forward when ready. Problem 50c

NOT:TISS:it is a person and it can fool everyone all the time. Bush is a person and Bush can fool everyone all the time. Bush is a person Bush can fool everyone all the time. Pb (y)(z)((Py & Tz)  Fbyz) FET y:FET z: If y is a person and z is a time then Bush can fool y at z (y)(z)((Py & Tz)  Fbyz) Let’s put the green together with an ampersand. Go forward when ready. Problem 50c

NOT:TISS:it is a person and it can fool everyone all the time. Bush is a person and Bush can fool everyone all the time. (Pb & (y)(z)((Py & Tz)  Fbyz)) Bush is a person Bush can fool everyone all the time. Pb (y)(z)((Py & Tz)  Fbyz) FET y:FET z: If y is a person and z is a time then Bush can fool y at z (y)(z)((Py & Tz)  Fbyz) The red came from the green “NOT:TISS:” statement, so let’s take Bush out and put a tilde on the outside. Go forward when ready. Problem 50c

NOT:TISS:it is a person and it can fool everyone all the time. ~(x)(Px & (y)(z)((Py & Tz)  Fxyz)) Bush is a person and Bush can fool everyone all the time. (Pb & (y)(z)((Py & Tz)  Fbyz)) Bush is a person Bush can fool everyone all the time. Pb (y)(z)((Py & Tz)  Fbyz) FET y:FET z: If y is a person and z is a time then Bush can fool y at z (y)(z)((Py & Tz)  Fbyz) The green says the same thing. Let’s do that one. Go forward when ready. Problem 50c

~(x)(Px & (y)(z)((Py & Tz)  Fxyz)) NOT:TISS:it is a person and it can fool everyone all the time. Bush is a person and Bush can fool everyone all the time. (Pb & (y)(z)((Py & Tz)  Fbyz)) Bush is a person Bush can fool everyone all the time. Pb (y)(z)((Py & Tz)  Fbyz) FET y:FET z: If y is a person and z is a time then Bush can fool y at z (y)(z)((Py & Tz)  Fbyz) Alternately, this could have been done as a universal statement. Let us see what that would look like in one slide. Go forward when ready. Problem 50c

~(x)(Px & (y)(z)((Py & Tz)  Fxyz)) (x)(Px  (y)(z)((Py & Tz) & ~Fxyz)) NOT:TISS:it is a person and it can fool everyone all the time. FET:If it is a person then there is at least one time and person such that the first person cannot fool the second person at that time. So there you have the three parts to problem 50. Let’s see how they fit together. Go forward when ready. Problem 50c

50a: You can fool some of the people all of the time.
You, (I.e. any person) can fool some of the people all of the time, and all of the people some of the time, but you cannot fool all of the people all of the time. 50a: You can fool some of the people all of the time. 50b:You can fool all of the people some of the time. 50c:You cannot fool all of the people all of the time. This breaks down to 50a, 50b and 50c, who’s representations are on the next page. Go forward when ready . Problem 50

50a: You can fool some of the people all of the time
You, (I.e. any person) can fool some of the people all of the time, and all of the people some of the time, but you cannot fool all of the people all of the time. 50a: You can fool some of the people all of the time (x)(Px (y)(Py & (z)(Tz  Fxyz))) 50b:You can fool all of the people some of the time (x)(Px (y)(Ty & (z)(Pz  Fxzy)) 50c:You cannot fool all of the people all of the time. ~(x)(Px & (y)(z)((Py & Tz)  Fxyz)) Now let’s put them all together for the final answer. Go forward when ready . Problem 50

50a: You can fool some of the people all of the time
You, (I.e. any person) can fool some of the people all of the time, and all of the people some of the time, but you cannot fool all of the people all of the time. (((x)(Px (y)(Py & (z)(Tz  Fxyz)))&(x)(Px (y)(Ty & (z)(Pz  Fxzy)))) &~(x)(Px & (y)(z)((Py & Tz)  Fxyz))) 50a: You can fool some of the people all of the time (x)(Px (y)(Py & (z)(Tz  Fxyz))) 50b:You can fool all of the people some of the time (x)(Px (y)(Ty & (z)(Pz  Fxzy)) 50c:You cannot fool all of the people all of the time. ~(x)(Px & (y)(z)((Py & Tz)  Fxyz)) So what this says then is that there is some person you can fool at all times and some time when you can fool all persons, but no one can fool all persons at all times. Go forward for new problem . . Problem 50

There is at least one thing that laughs at death.
New Problem Let the dictionary be Lxy: x laughs at y and d:death. This problem will not involve identity and is easy but it will be important to contrast it to later problems Go forward when ready to put in ‘TISS’ standard form. Problem 1 with identity

There is at least one thing that laughs at death.
TISS: it laughs at death. Easy enough, Let’s symbolize the red and thus the green. Go forward when ready to put in ‘TISS’ standard form. Problem 1 with identity

There is at least one thing that laughs at death. (x)Lxd
TISS: it laughs at death. . Go forward when ready for new problem. Problem 1 with identity

There is at most one thing that laughs at death.
New Problem This one will use identity and it involves a trick that might not be apparent at first. To say that at most one thing has a certain property means no more than one. So that would be 1 or zero. So we cannot start with a ‘TISS’ statement because that gives us at least one thing. To say that there is a most one thing says that if you are thinking about two things like that then they are really one thing. How would we do that in standard form? Go forward when ready for new problem. Problem 2 with identity

FET x:FET y:if x laughs at death and y laughs at death then x is really y. We should be able to turn the red and the green into symbols now, Go forward when ready for new problem. Problem 2 with identity

(x)(y)((Lxd & Lyd)  x=y) FET x:FET y:if x laughs at death and y laughs at death then x is y. The strange thing about this statement is that although it seems to be talking about one thing, it involves two quantifiers. That’s because you can also take this statement to be saying there are not 2 or more things. Although it is logically equivalent and thus could be represented the same way, by sticking close to the syntax of English, for the next problem, let us see what the symbolization of ‘not two or more’ would look like. Go forward when ready for new problem. Problem 2 with identity

It is not true that at least two things laugh at death.
New Problem Clearly this is a ‘NOT’ statement. What does it deny? Go forward when ready for ‘NOT’ standard form. Problem 3 with identity

NOT: There are at least two things that laugh at death. Let’s peel the ‘NOT’ off for simplicity. Go forward when ready. Problem 3 with identity

NOT: There are at least two things that laugh at death. There are at least two things that laugh at death To say ‘there are at least two things’ sounds like a double ‘TISS’ statement. So let us start it as ‘TISS x: TISS y: x laughs at death and y laughs at death… But this is not enough yet, because x and y could be the same thing. So how should we finish? Go forward when ready. Problem 3 with identity

NOT: There are at least two things that laugh at death. There are at least two things that laugh at death TISS x:TISS y: x laughs at death and y laughs at death and x is not identical to y. We should be able to turn the red and thus the green into symbols now. Go forward when ready. Problem 3 with identity

NOT: There are at least two things that laugh at death. There are at least two things that laugh at death (x)(y)((Lxd & Lyd) & ~x=y) TISS x:TISS y: x laughs at death and y laughs at death and x is not identical to y. The green is just “NOT’ red and the brown is synonymous. So let us finish this. Go forward when ready. Problem 3 with identity

~(x)(y)((Lxd & Lyd) & ~x=y) NOT: There are at least two things that laugh at death. There are at least two things that laugh at death (x)(y)((Lxd & Lyd) & ~x=y) TISS x:TISS y: x laughs at death and y laughs at death and x is not identical to y. This was an equivalent way of saying problem 2. So let’s see them together. Go forward when ready. Problem 3 with identity

(x)(y)((Lxd & Lyd)  x=y) It is not true that at least two things laugh at death. ~(x)(y)((Lxd & Lyd) & ~x=y) It is always the case that when we have a universal representation, that there is a tilde existential representation that is logically equivalent to it. Go forward when ready for new problem. Problem 2 and 3 with identity

There is exactly one thing that laughs at death.
New Problem This is the content of problem 1 and problem 2 thrown together, so we could just throw an ampersand between them. But there is a way to treat them as an existential. State it as a ‘TISS’ statement. Go forward when ready. Problem 4 with identity

TISS: it laughs at death and anything that laughs at death is it. Let’s pick a person, like Alan Go forward when ready. Problem 4 with identity

TISS: it laughs at death and anything that laughs at death is it. Alan laughs at death and anything that laughs at death is Alan. We should be able to symbolize this now. Give it a try. Imagine and then go forward. Problem 4 with identity

TISS: it laughs at death and anything that laughs at death is it. Alan laughs at death and anything that laughs at death is Alan. (Lad & (y)(Lyd  y=a)) Let’s build the green statement, which is synonymous with the brown. Imagine and then go forward. Problem 4 with identity

(x)(Lxd & (y)(Lyd  y=x)) TISS: it laughs at death and anything that laughs at death is it. Alan laughs at death and anything that laughs at death is Alan. (Lad & (y)(Lyd  y=a)) Now let’s compare this to 1 and 2 on the next page Go forward when ready. Problem 4 with identity

(x)(Lxd & (y)(Lyd  y=x)) There is at least one thing that laughs at death. (x)Lxd There is at most one thing that laughs at death. (x)(y)((Lxd & Lyd)  x=y) The brown is a little simpler than just sticking an ampersand between the green and red. ‘Exactly’ statements will have a form that looks something like this. We will see more. Go forward when ready for new problem. Problem 4, 1 and 2 with identity

There are at least two things that laugh at death.
New Problem We did this before as part of problem 3. Try to remember and turn it into symbols before seeing next slide. Go forward when ready. Problem 5 with identity

There are at least two things that laugh at death.
(x)(y)((Lxd & Lyd) & ~x=y) Very good. Go forward when ready for new problem.. Problem 5 with identity

There are at most two things that laugh at death.
New Problem We did ‘there is at most one thing..’. How will this be different? ‘At most one thing…’ took 2 quantifiers, so if you guessed that this will take 3 you are correct. Start it with a triple FET Go forward when ready. Problem 6 with identity

FET x: FET y: FET z: if x laughs at death and y laughs at death and z laughs at death then either x equals y or y equals z or x equals z. In other words. If you think you have 3 things that laugh at death, at least one of them is a repeat. Let’s turn it into symbols on the next slide Go forward when ready. Problem 6 with identity

FET x: FET y: FET z: if x laughs at death and y laughs at death and z laughs at death then either x equals y or y equals z or x equals z. (x)(y)(z)(((Lxd & Lyd) & Lzd)  ((x=y v y=z) v x=z)) So the green should be represented the same way. Go forward when ready. Problem 6 with identity

(x)(y)(z)(((Lxd & Lyd) & Lzd)  ((x=y v y=z) v x=z)) FET x: FET y: FET z: if x laughs at death and y laughs at death and z laughs at death then either x equals y or y equals z or x equals z. Go forward when ready for a new problem. Problem 6 with identity

There are exactly two things that laugh at death.
New Problem Let us start this as a double ‘TISS’ How will it finish? Go forward when ready for part of a standard form. Problem 7 with identity

TISS x:TISS y: x laughs at death, y laughs at death, x and y are not identical ……. This gives us at least two things. How will we say that there are no more? Go forward when ready to finish standard form. Problem 7 with identity

TISS x:TISS y: x laughs at death, y laughs at death, x and y are not identical and anything that laughs at death is identical to x or y. The red part tells us that there are no more laughing at death things than the ones we already have. Now let’s symbolize on the next page. Go forward when ready for symbols. Problem 7 with identity

(x)(y)(((Lxd & Lyd) & ~x=y) & (z)(Lzd  (z=x v z=y))) TISS x:TISS y: x laughs at death, y laughs at death, x and y are not identical and anything that laughs at death is identical to x or y. Now let’s contrast ‘at least two things.’., ‘at most two things..’ and ‘exactly two things..’ on the next page. Go forward when ready. Problem 7 with identity

(x)(y)(((Lxd & Lyd) & ~x=y) & (z)(Lzd  (z=x v z=y))) There are at least two things that laugh at death. (x)(y)((Lxd & Lyd) & ~x=y) There are at most two things that laugh at death. (x)(y)(z)(((Lxd & Lyd) & Lzd)  ((x=y v y=z) v x=z)) How does this compare to our statements about one thing? Go forward when ready. Problem 5, 6 and 7 with identity

(x)(Lxd & (y)(Lyd  y=x)) There is at least one thing that laughs at death. (x)Lxd There is at most one thing that laughs at death. (x)(y)((Lxd & Lyd)  x=y) There are exactly two things that laugh at death. (x)(y)(((Lxd & Lyd) & ~x=y) & (z)(Lzd  (z=x v z=y))) There are at least two things that laugh at death. (x)(y)((Lxd & Lyd) & ~x=y) There are at most two things that laugh at death. (x)(y)(z)(((Lxd & Lyd) & Lzd)  ((x=y v y=z) v x=z)) Now you may be able to see the pattern for doing any number of things. Of course as the number increases the formula will become geometrically more complex. Go forward when ready for new problem. Comparison of 1, 2, 3, 5, 6, 7

There are at most three things that laugh at death
New Problem We’ll look at this one to see what’s involved when numbers get bigger. We know that the ‘at most’ problems are always synonymous with a FET standard form, so let us begin that way. Note that ‘There are at most two…’ started with 3 FETs.. So you can imagine how many this will start with. Go forward when ready. Problem 8 with identity

There are at most three things that laugh at death.
FET x:FET y: FET z: FET x1: if x laughs at death and y laughs at death and z laughs at death and x1 laughs at death then…. How will we finish this? If we have picked 4 things that laugh at death and there are at most 3 things then what do we know? Go forward when ready. Problem 8 with identity

FET x:FET y: FET z: FET x1: if x laughs at death and y laughs at death and z laughs at death and x1 laughs at death then either x equals y or x equals z or x equals x1 or y equals z or y equals x1 or z equals x1. And this clearly shows the pattern and shows that the length of the equation will get progressively larger as the number increases. Let’s turn it into symbols. Go forward when ready for symbols. Problem 8 with identity

FET x:FET y: FET z: FET x1:If x laughs at death and y laughs at death and z laughs at death and x1 laughs at death then either x equals y, or x equals z, or x equals x1, or y equals z, or y equals x1, or z equals x1. (x)(y)(z)(z1)((((Lxd & Lyd) & Lzd) & Lx1d)  (x=y v (x=z v (x=x1 v (y=z v (y=x1 v z=x1)))))) And so you see why higher numbers make things more complex. And of course, the green being synonymous would be represented the same way. Go forward when ready for new problem. Problem 8 with identity

There are exactly two integers between 3 and 6.
New Problem We can start the ‘exactly’ problems with ‘TISS’. We have to say there are at least two and no more. Go forward when ready for standard form.. Problem 10 with identity

TISS x:TISS y: x is and integer, y is an integer, x is between 3 and 6 and y is between 3 and 6 and x and y are not identical and anything that is between 3 and 6 is identical to x or y. For representation let t:three, s:six, Ix:x is an integer and Bxyz: x is between y and z Go forward when ready for representation. Problem 10 with identity

TISS x:TISS y: x is and integer, y is an integer, x is between 3 and 6 and y is between 3 and 6 and x and y are not identical and anything that is between 3 and 6 is identical to x or y. (x)(y)((((Ix & Iy)& (Bxts & Byts)) & ~x=y) & (z)(Bzts  (z=x or z=y))) And of course the green gets represented the same way. Go forward when ready for new set of problems. Problem 10 with identity

Go forward when ready for new problem.
For the rest of the identity problems, let’s use the same dictionary for all. Let your domain be the counting numbers (1,2,3), so you do not have to say x is a positive integer. Let a:1, b:2, c:3, d:4 e:5, f:6, Ex: x is even, Px: x is prime, Lxy: x is less than y, Txyz: x times y equals z, Sxyz: x plus y equals z. Go forward when ready for new problem. Problem 10 with identity

Go forward when ready for standard form.
2 is the only even prime. New Problem For this what do we have to say about two? That it is an even prime and there are no others. Go forward when ready for standard form. Problem 11 with identity

Go forward when ready for symbols.
2 is the only even prime. Two is even and two is prime and anything that is both even and prime is identical to two. Let’s try symbols now. Go forward when ready for symbols. Problem 11 with identity

((Eb & Pb) & (x)((Ex & Px) x=b))
2 is the only even prime. Two is even and two is prime and anything that is both even and prime is identical to two. ((Eb & Pb) & (x)((Ex & Px) x=b)) And green is represented the same way. Go forward when ready for new problem. Problem 11 with identity

The only even prime is less than 3.
New Problem How would we state this in standard form? We would say that there is an even prime less than three and all even primes less than three are identical to it. Go forward when ready for standard form. Problem 12 with identity

TISS: It is an even prime and every even prime is identical to it, and it is less than three. Now let’s turn it into symbols. Go forward when ready for standard form. Problem 12 with identity

TISS: It is an even prime and every even prime is identical to it, and it is less than three. (x)(((Ex & Px) & (y)((Ex & Px)  y=x)) & Lxc) Another way to do the same thing would be to say there is an even prime and anything not identical to it is not an even prime.and it is less than three. Go forward when ready for that approach. Problem 12 with identity

TISS: It is an even prime and every even prime is identical to it. (x)(((Ex & Px) & (y)((Ex & Px)  y=x)) & Lxc) TISS: It is an even prime and everything not identical to it is not an even prime and it is less than three. (x)(((Ex & Px) & (y)(~y=x  ~(Ex & Px))) & Lxc) These are logically equivalent and thus either would be an acceptable answer for the brown Go forward when ready for new problem.. Problem 12 with identity

Every two numbers have one and only one sum. New Problem
This will come in handy since many of the problems to come talk about THE sum of two numbers. What does that mean? That there is at least one sum of any two numbers and no more. Go forward when ready for standard form Unique sum with identity

Every two numbers have one and only one sum.
FET x:FET y: there is one and only one sum of x and y. Let’s imagine a couple of numbers Go forward when ready for plug in Unique sum with identity

FET x:FET y: there is one and only one sum of x and y. There is one and only one sum of 3 and 4. But how would we state this in standard form? Go forward when ready for plug in Unique sum with identity

FET x:FET y: there is one and only one sum of x and y. There is one and only one sum of 3 and 4. TISS: it is the sum of 3 and 4 and anything that is the sum of 3 and 4 is identical to it. But how would we state this in standard form? Go forward when ready for plug in Unique sum with identity

FET x:FET y: there is one and only one sum of x and y. There is one and only one sum of 3 and 4. TISS: it is the sum of 3 and 4 and anything that is the sum of 3 and 4 is identical to it. Now we can turn this into symbols, right? Go forward when ready for symbolization. Unique sum with identity

FET x:FET y: there is one and only one sum of x and y. There is one and only one sum of 3 and 4. TISS: it is the sum of 3 and 4 and anything that is the sum of 3 and 4 is identical to it. (z)(Scdz & (x1)(Scdx1  x1=z)) This would be the same as the green. So let’s do the green. Go forward when ready for symbolization of green. Unique sum with identity

FET x:FET y: there is one and only one sum of x and y. There is one and only one sum of 3 and 4. (z)(Scdz & (x1)(Scdx1  x1=z)) TISS: it is the sum of 3 and 4 and anything that is the sum of 3 and 4 is identical to it. (x)(Scdx & (y)(Scdy  y=x)) But the red just uses 3 and 4 as examples, so we could just generalize on c and d to get the green. Go forward when ready for symbolization. Unique sum with identity

FET x:FET y: there is one and only one sum of x and y. (x)(y)(z)(Sxyz & (x1)(Sxyx1  x1=z)) There is one and only one sum of 3 and 4. (x)(Scdx & (y)(Scdy  y=x)) TISS: it is the sum of 3 and 4 and anything that is the sum of 3 and 4 is identical to it. And of course, the green will be the same Go forward when ready for new problem. Unique sum with identity

Every two numbers have one and only one product. New Problem
Gosh, we just did a problem like this (look back). So let us jump to the symbolization Go forward when ready for answer. Unique product with identity

Every two numbers have one and only one product.
(x)(y)(z)(Txyz & (x1)(Txyx1  x1=z)) Keep these in mind, when doing the rest of the problems. Go forward when ready for answer. Unique product with identity

There is exactly one number whose square is less than 2.
New Problem Let’s state it in standard form. We know it should start with a ‘TISS’ Go forward when ready for new standard form. Problem 15 with identity

TISS: the product of it times itself is less than 2 and any number whose square is less than 2 is identical to it. Let’s try plugging in a term, like 1 for example. Go forward when ready for new standard form. Problem 15 with identity

TISS: the product of it times itself is less than 2 and any number whose square is less than 2 is identical to it. The product of 1 times 1 is less than 2 and any number whose square is less than 2 is identical to 1. At this point we have to decide how to treat ‘the product’. Should we say ‘there is a product...’ or ‘every product…’? It turns out it doesn’t matter if we tag ‘(x)(y)(z)(Txyz & (x1)(Txyx1  x1=z))’on the end of the formula, since it talks about ‘THE’ product. So let’s imagine it on the end whether we actually put it there or not and treat it as if it said ‘some product’. Let’s imagine the standard form Go forward when ready for new standard form. Problem 15 with identity

TISS: the product of it times itself is less than 2 and any number whose square is less than 2 is identical to it. The product of 1 times 1 is less than 2 and any number whose square is less than 2 is identical to that product. TISS: it is the product of 1 times 1 and it is less than 2 and any number whose square is less than 2 is identical to it. Let’s try symbols now Imagine and go forward when ready for symbols. Problem 15 with identity

TISS: the product of it times itself is less than 2 and any number whose square is less than 2 is identical to it. The product of 1 times 1 is less than 2 and any number whose square is less than 2 is identical to 1. TISS: it is the product of 1 times 1 and it is less than 2 and any number whose square is less than 2 is identical to 1. (x)((Taax & Lxb) & (y)(x1)((Tx1x1y & Lyb)  x1=a)) This says the same as the green, so let’s do the green. Imagine and go forward when ready for symbols of green. Problem 15 with identity

TISS: the product of it times itself is less than 2 and any number whose square is less than 2 is identical to it. The product of 1 times 1 is less than 2 and any number whose square is less than 2 is identical to 1. (x)((Taax & Lxb) & (y)(x1)((Tx1x1y & Lyb)  x1=a)) TISS: it is the product of 1 times 1 and it is less than 2 and any number whose square is less than 2 is identical to 1. One was an example of the something. So let’s represent the green. Imagine and go forward when ready for symbols of green. Problem 15 with identity

TISS: the product of it times itself is less than 2 and any number whose square is less than 2 is identical to it. (z)(x)((Tzzx & Lxb) & (y)(x1)((Tx1x1y & Lyb)  x1=z)) The product of 1 times 1 is less than 2 and any number whose square is less than 2 is identical to 1. (x)((Taax & Lxb) & (y)(x1)((Tx1x1y & Lyb)  x1=a)) TISS: it is the product of 1 times 1 and it is less than 2 and any number whose square is less than 2 is identical 1. And of course the green is synonymous and thus represent the same way. But if we want to make sure that there is one and only one product, we would have to add on to it ‘(x)((Taax & Lxb) & (y)(x1)((Tx1x1y & Lyb)  y=x))’ Imagine and go forward when ready for new problem. Problem 15 with identity

New Problem The smallest even number is greater than 1.
To say that something is the smallest even number is to say that no even number is less than that. So let us first say there there is a smallest even number and then say it is greater than 1. Imagine standard form and then go forward. Problem 16 with identity

The smallest even number is greater than 1.
TISS: it is even and it is less than every other even number and it is greater than 1 So let us plug something in and see how it goes. Imagine standard form and then go forward. Problem 16 with identity

TISS: it is even and it is less than every other even number and it is greater than 1 2 is even and 2 is less than every other even number and 2 is greater than 1. This is an and statement, maybe two ‘and’s so let’s break these into 3 parts and see how it goes. Imagine standard form and then go forward. Problem 16 with identity

TISS: it is even and it is less than every other even number and it is greater than 1 2 is even and 2 is less than every other even number and 2 is greater than 1. This is an ‘and’ statement, maybe two ‘and’s so let’s break these into 3 parts and see how it goes. (This should be O.K. because to put a double ‘and’ statement together will just take a couple of ampersands and it doesn’t matter where the parentheses go.) Imagine break up and then go forward. Problem 16 with identity

TISS: it is even and it is less than every other even number and it is greater than 1 2 is even and 2 is less than every other even number and 2 is greater than 1. 2 is even is less than every other even number is greater than 1. The left and the right should not be that hard. Let’s symbolize those. Imagine symbols and then go forward. Problem 16 with identity

TISS: it is even and it is less than every other even number and it is greater than 1 2 is even and 2 is less than every other even number and 2 is greater than 1. 2 is even is less than every other even number is greater than 1. Eb Lab But to represent the middle red is more complex. How will we say that 2 is less than every OTHER even number? What would that standard form look like? Imagine standard form and then go forward. Problem 16 with identity

FET:if it is an even number not identical to 2 then 2 is less than it.
The smallest even number is greater than 1. TISS: it is even and it is less than every other even number and it is greater than 1 2 is even and 2 is less than every other even number and 2 is greater than 1. 2 is even is less than every other even number is greater than 1. Eb Lab FET:if it is an even number not identical to 2 then 2 is less than it. O.K. We can turn that into symbols. Imagine symbols and then go forward. Problem 16 with identity

FET:if it is an even number not identical to 2 then 2 is less than it.
The smallest even number is greater than 1. TISS: it is even and it is less than every other even number and it is greater than 1 2 is even and 2 is less than every other even number and 2 is greater than 1. 2 is even is less than every other even number is greater than 1. Eb Lab FET:if it is an even number not identical to 2 then 2 is less than it. (x)((Eb & ~x=b)  Lbx) So that is the same as the green, since the green is synonymous with the red English. Imagine and then go forward. Problem 16 with identity

Eb (x)((Eb & ~x=b)  Lbx) Lab
The smallest even number is greater than 1. TISS: it is even and it is less than every other even number and it is greater than 1 2 is even and 2 is less than every other even number and 2 is greater than 1. 2 is even is less than every other even number is greater than 1. Eb (x)((Eb & ~x=b)  Lbx) Lab FET:if it is an even number not identical to 2 then 2 is less than it. (x)((Eb & ~x=b)  Lbx) The green is a double ‘and’ statement so let’s put it together with the red formulas. The position of the parentheses don’t matter on this one, as long as we make it a statement. Imagine and then go forward. Problem 16 with identity

((Eb & (x)((Eb & ~x=b)  Lbx)) & Lab)
The smallest even number is greater than 1. TISS: it is even and it is less than every other even number and it is greater than 1 2 is even and 2 is less than every other even number and 2 is greater than 1. ((Eb & (x)((Eb & ~x=b)  Lbx)) & Lab) 2 is even is less than every other even number is greater than 1. Eb (x)((Eb & ~x=b)  Lbx) Lab FET:if it is an even number not identical to 2 then 2 is less than it. (x)((Eb & ~x=b)  Lbx) But ‘2’ was just an example of the ‘TISS’ green statement. So let’s represent the green.. Imagine and then go forward. Problem 16 with identity

(y)((Ey & (x)((Ey & ~x=y)  Lyx)) & Lay)
The smallest even number is greater than 1. TISS: it is even and it is less than every other even number and it is greater than 1 (y)((Ey & (x)((Ey & ~x=y)  Lyx)) & Lay) 2 is even and 2 is less than every other even number and 2 is greater than 1. ((Eb & (x)((Eb & ~x=b)  Lbx)) & Lab) 2 is even is less than every other even number is greater than 1. Eb (x)((Eb & ~x=b)  Lbx) Lab FET:if it is an even number not identical to 2 then 2 is less than it. (x)((Eb & ~x=b)  Lbx) And the red is synonymous with the green and so, of course, the green should be represented the same way. Go forward when ready for new problem. Problem 16 with identity

There are at least 3 prime numbers that are less than 6.
New Problem This one should not be too hard. ‘At least’ always seems a bit easier than ‘at most’ or exactly. Let’s start with a triple TISS statement. Go forward for start of standard form. Problem 17 with identity

TISS x:TISS y:TISS z: x is prime and less than 6 and y is prime and less than 6 and z is prime and less than 6……... Does this give us 3 statements? No. We need to also say that they are different from each other Go forward when ready to finish standard form. Problem 17 with identity

TISS x:TISS y:TISS z: x is prime and less than 6 and y is prime and less than 6 and z is prime and less than 6 and x is not identical to y and y is not identical to z and x is not identical to z. We should be able to turn this into symbols now. Go forward when ready for symbols. Problem 17 with identity

TISS x:TISS y:TISS z: x is prime and less than 6 and y is prime and less than 6 and z is prime and less than 6 and x is not identical to y and y is not identical to z and x is not identical to z. (x)(y)(z)((((((Px & Lxf)& (Py & Lyf))& (Pz & Lzf)) & ~x=y) & ~y=z) & ~x=z) And it will be the same formula for the green. Go forward when ready for new problem. Problem 17 with identity

One is the only number that is its own square root. New Problem
To say that one is the only number that is its own square root says that it is it’s own square root and no other number is its own square root. So what will the standard form look like? Go forward when ready for standard form. Problem 18 with identity

One is the only number that is its own square root.
One times itself is one and anything that times itself is itself is one. Let’s try symbols on the next step. Go forward when ready for symbols. Problem 18 with identity

One is the only number that is its own square root.
One times itself is one and anything that times itself is itself is one. (Taaa & (x)(Txxx  x=a)) And the green would be the same. If we wanted to capture the notion that every number has at most one square root. We could tack that on. Let’s see what that would look like next. Go forward when ready for new problem. Problem 18 with identity

Every number has at most one square root. New Problem
You might object to the truth of this statement, since the square root of 4 is both +2 and -2. The use of ‘THE’ square root seems inappropriate. However, remember that we are only using the counting numbers starting from one. Besides we don’t have to only symbolize true statements. Should we have said ‘Every number has exactly one square root’? Again, noting that our domain is the counting numbers some of them do not have square roots. So if we wanted to symbolize a true and strong statement in our domain, this is about as strong a statement as we can make about the existence of square roots and the counting numbers. Let’s put it into standard form. Go forward when ready for standard form. Unique square root with identity

Every number has at most one square root.
FET x: there is at most one number y such that y times y equals x. Let’s try symbols now. Go forward when ready for symbols. Unique square root with identity

Every number has at most one square root.
FET x: there is at most one number y such that y times y equals x. (x)(y)(z)((Tyyx & Tzzx)  y=z)) And the green would be the same Go forward when ready for new problem. Unique square root with identity

There are exactly 3 odd numbers that are less than 6. New Problem
This says there are three odd numbers and at most three odd numbers. Go forward when ready to put in standard form. Problem 19 with identity

There are exactly 3 odd numbers that are less than 6.
TISS x:TISS y:TISS z: x is odd and less than 6, y is odd and less than 6 and z is odd and less than 6 and x is not equal to y and y is not equal to z and x is not equal to z and anything that is odd and less than 6 is identical to x or y or z. So let’s turn it into symbols in the next slide Go forward when ready to see symbols. Problem 19 with identity

There are exactly 3 odd numbers that are less than 6.
TISS x:TISS y:TISS z: x is odd and less than 6, y is odd and less than 6 and z is odd and less than 6 and x is not equal to y and y is not equal to z and x is not equal to z and anything that is odd and less than 6 is identical to x or y or z. (x)(y)(z)((((((~Ex&~Ey)&~Ez)&Lxf) & Lyf) & Lzf) & (x1)((~Ex1 & Lx1f)  ((x1=x v x1=y) v x1=z))) And that is going to be the same answer for the green. Notice our dictionary had ‘Even’ as a predicate and we capture the concept of ‘odd’ above by just saying it is not even. Go forward when ready conclude. Problem 19 with identity

Dubadubaduba That’s All Folks The end of the cartoon.

Important note. All references to ‘For At Least One Thing:’ FALOT’ have been replaced with ‘There is something such that:’ ‘TISS’ Important Change.

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Important note. All references to ‘For At Least One Thing:’ FALOT’ have been replaced with ‘There is something such that:’ ‘TISS’ Important Change.

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Presentation on theme: "Important note. All references to ‘For At Least One Thing:’ FALOT’ have been replaced with ‘There is something such that:’ ‘TISS’ Important Change."— Presentation transcript:

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