Presentation on theme: "Exam #3 will be given on Monday Nongraded Homework: Now that we are familiar with the universal quantifier, try"— Presentation transcript:
Exam #3 will be given on Monday Nongraded Homework: Now that we are familiar with the universal quantifier, try http://www.poweroflogic.com/cgi/menu.cgi http://www.poweroflogic.com/cgi/menu.cgi (9.1, C, D, E and F – see ‘help’ link for symbol use; note about upside-down ‘A’)
Exam #3 One proof without SI or TI rules (10 pts.) Two proofs using any of our rules (6 and 10 pts.) Four symbolizations in LMPL (three, three, four, and four points)
The Universal Quantifier ( x), ( y), ( z), using other small-case letters after ‘s’, if needed. It means, “the following is true of every single thing in the universe” or “for every value of x, the following comes out true” Like the existential quantifier and the tilde, the universal quantifier applies to (or has in its scope) whatever immediately follows it.
Guidelines for symbolization in LMPL: 1. When using the universal quantifier in translation, use an arrow as the main operator within the scope of the quantifier (almost always). All philosophers are mortal: ( x)(Px → Mx) In other words, for every single thing in the universe, if it’s a philosopher, then it’s mortal.
No trees are animals. ( x)(Tx → ~ Ax) The group you’re talking about in its entirety is the group of trees. What does ‘( x)(Tx & ~ Ax)’ say? For anything in the universe, it is a tree and not an animal. In other words, everything in the universe is a tree and not an animal. This formula is well formed and it has truth-conditions, but it’s unlikely to appear in a symbolization, because it’s unlikely anyone would ever say something with that formula’s truth-conditions.
Exceptions What about, “Everything has mass and charge”? ( x)(Mx & Cx) is the right translation, but this is a very rare case where the speaker wants to say that both predicates apply to every single thing in the universe.
B. All people are happy. Looks like it should be ( x)(Px → Hx) But since Forbes allows persons to be a domain of discourse, we don’t actually need ‘Px’ or the arrow ( x)Hx is fine. Let the dictionary be your guide on HW and exams. If ‘P_: _ is a person’ appears in the dictionary, use it in your symbolization.
2. When using the existential quantifier, an ampersand should be the main connective within the scope of the quantifier (almost always). Some philosophers are happy. (P_: _ is a philosopher; H_: _ is happy) ( x)(Px & Hx)
Why not ( x)(Px → Hx)? It’s well formed, but it doesn’t have the right truth-conditions. It says, “for at least one thing P_ → H_ is true” (with the same thing put in both blanks). But it’s too easy to make P_ → H_ true. Just find something that isn’t a philosopher (use a table, for example). F → [any truth-value] is true.
Exceptions? What about, “Something is either rotten or dead”? ( x)(Rx v Dx) is the right answer, but this is a rare case. Even here, the person speaking would probably mean “Something in here is either rotten or dead.” How would you translate that?
3. When using the universal quantifier in translation, put the group you wish to talk about in its entirety to the left of the arrow that is the m.c. inside the scope of the universal quantifier. All frogs are green. ( x)(Fx → Gx) ‘Fx’ is the antecedent because we want to say something about all frogs, not about all green things.
What about this one? Only U.S. citizens are allowed to be president. (Cx: x is a U.S. citizen; Ax: x is allowed to be president) ( x)(Ax → Cx) We want to say something about all people allowed to be president, not about all U.S. citizens.
4. When translating, group a noun and its modifier together around an ampersand. All green frogs are poisonous. ( x)[(Gx & Fx) → Px] All frogs from Brazil are poisonous. ( x)[(Fx & Bx) → Px]
Every philosopher who lives in Brazil speaks German. ( x)[(Px & Bx) → Sx] All frogs are slimy amphibians. ( x)[Fx → (Sx & Ax)] Some rich philosophers are humble. ( x)[(Rx & Px) & Hx]
When should a quantifier not be the main operator of a predicate logic symbolization? Some philosophers are good, and some philosophers are not good. ( x)(Px & Gx) & ( x)(Px & ~ Gx) There are two quantifiers, but neither is the main connective.
Guidelines for deciding whether the initial quantifier should be the main operator: 1. Apply the sandwich principle: Everything between an associated ‘either’ and ‘or’ or ‘if’ and ‘then’ should be grouped together. If any witness told the truth, then George is guilty. (W_: _ is a witness; T_: _ told the truth; G_: _ is guilty; g: George)
( x)(Wx & Tx) → Gg Either all of the witnesses told the truth, or George is guilty. (G_: _ is guilty; T_: _ told the truth; W_: _ is a witness; g: George) ( x)(Wx → Tx) v Gg
2. BUT, if, each time you pick a value for x, you want to talk about the same thing throughout the entire instance, the initial quantifier should be the m.c. If any witness told the truth, then he or she is honest. (T_: _ told the truth; W_: _ is a witness; H_: _ is honest) ( x)[(Wx & Tx) → Hx]
Speakers have their names listed in the program only if they are famous. (Sx: x is a speaker; Px: x’s name is listed in the program; Fx: x is famous) ( x)[(Sx & Px) → Fx]
Some experienced mechanics are well paid only if all the inexperienced ones are lazy. (E_: _ is experienced; M_: _ is a mechanic; W_: _ is well paid; L_: _ is lazy) ( x)[(Ex & Mx) & Wx] → ( x)[(Mx & ~Ex) → Lx ]
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