# Semantic Paradoxes. THE BARBER The Barber Paradox Once upon a time there was a village, and in this village lived a barber named B.

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THE BARBER

The Barber Paradox Once upon a time there was a village, and in this village lived a barber named B.

The Barber Paradox B shaved all the villagers who did not shave themselves, And B shaved none of the villagers who did shave themselves.

The Barber Paradox Question, did B shave B, or not?

Suppose B Shaved B 1. B shaved BAssumption 2. B did not shave any villager X where X shaved X Assumption 3. B did not shave B1,2 Logic

Suppose B Did Not Shave B 1. B did not shave BAssumption 2. B shaved every villager X where X did not shave X Assumption 3. B shaved B1,2 Logic

Contradictions with Assumptions We can derive a contradiction from the assumption that B shaved B. We can derive a contradiction from the assumption that B did not shave B.

The Law of Excluded Middle Everything is either true or not true. Either P or not-P, for any P. Either B shaved B or B did not shave B, there is not third option.

It’s the Law Either it’s Tuesday or it’s not Tuesday. Either it’s Wednesday or it’s not Wednesday. Either killing babies is good or killing babies is not good. Either this sandwich is good or it is not good.

Disjunction Elimination A or B A implies C B implies C Therefore, C

Example Either Michael is dead or he has no legs If Michael is dead, he can’t run the race. If Michael has no legs, he can’t run the race. Therefore, Michael can’t run the race.

Contradiction, No Assumptions B shaves B or B does not shave B [Law of Excluded Middle] If B shaves B, contradiction. If B does not shave B, contradiction. Therefore, contradiction

Contradictions Whenever we are confronted with a contradiction, we need to give up something that led us into the contradiction.

Give up Logic? For example, we used Logic in the proof that B shaved B if and only if B did not shave B. So we might consider giving up logic. A or B A implies C B implies C Therefore, C

No Barber In this instance, however, it makes more sense to give up our initial acquiescence to the story: We assumed that there was a village with a barber who shaved all and only the villagers who did not shave themselves.

The Barber Paradox The paradox shows us that there is no such barber, and that there cannot be.

Semantic Paradoxes Unfortunately, much of our semantic vocabulary like ‘is true’ and ‘applies to’ leads us into contradictions where it is highly non-obvious what to abandon.

Disquotation To say P is the same thing as saying ‘P’ is true. This is the “disquotation principle”: P = ‘P’ is true

Liar Sentence The liar sentence is a sentence that says that it is false. For example, “This sentence is false,” or “The second example sentence in the powerpoint slide titled ‘Liar Sentence’ is false.”

Liar Sentence L = ‘L’ is not true

“‘L’ is true” 1. ‘L’ is trueAssumption 2. L 1, Disquotation 3. ‘L’ is not true2, Def of L 1 & 3 form a contradiction

“‘L’ is not true” 1. ‘L’ is not trueAssumption 2. L1, Def of L 3. ‘L’ is true2, Disquotation 1 & 3 form a contradiction

Contradiction Thus we can derive a contradiction from the assumption that “‘L’ is true or ‘L’ is not true,” [Law of Excluded Middle] plus the inference rule: A or B A implies C B implies C Therefore, C

Contradiction ‘L’ is true or ‘L’ is not true [Law of Excluded Middle] If ‘L’ is true, then ‘L’ is true and not true. If ‘L’ is not true, then ‘L’ is true and not true. Therefore, ‘L’ is true and not true.

Solutions 1.Give up excluded middle 2.Give up disjunction elimination 3.Give up disquotation 4.Disallow self-reference 5.Accept that some contradictions are true

1. Giving up Excluded Middle The problem with giving up the Law of Excluded Middle is that it seems to collapse into endorsing contradictions: “According to LEM, every sentence is either true or not true. I disagree: I think that some sentences are not true and not not true at the same time.”

2. Give up Disjunction Elimination Basic logical principles are difficult to deny. What would a counterexample to disjunction elimination look like? A or B A implies C B implies C However, not-C

3. Give up Disquotation Principle Giving up the disquotation principle P = ‘P’ is true Involves accepting that sometimes P but ‘P’ is not true or accepting that not-P but ‘P’ is true.

4. Disallow Self-Reference The problem with disallowing self-reference is that self-reference isn’t essential to the paradox. A: ‘B’ is true B: ‘A’ is not true

Circular Reference A B ‘B’ is true. ‘A’ is false.

Assume ‘A’ Is True A B ‘B’ is true. ‘A’ is false.

Then ‘B’ Is Also True A B ‘B’ is true. ‘A’ is false.

But Then ‘A’ is False! A B ‘B’ is true. ‘A’ is false.

Assume ‘A’ Is False A B ‘B’ is true. ‘A’ is false.

Then ‘B’ Is Also False A B ‘B’ is true. ‘A’ is false.

But Then ‘A’ Is Also True A B ‘B’ is true. ‘A’ is false.

“‘A’ is true” 1. ‘A’ is trueAssumption 2. A1, Disquotation 3. ‘B’ is true2, Def of A 4. B3, Disquotation 5. ‘A’ is not true4, Def of B

“‘A’ is not true” 1. ‘A’ is not trueAssumption 2. B1, Def of B 3. ‘B’ is true2, Disquotation 4. A3, Def of A 5. ‘A’ is true4, Disquotation

Contradiction, No Assumptions Either ‘A’ is true or ‘A’ is not true. [Law of Excluded Middle] If ‘A’ is true, then ‘A’ is true and not true. If ‘A’ is not true, then ‘A’ is true and not true. Therefore, ‘A’ is true and not true.

Disallowing Circular Reference Even circular reference is not essential. Stephen Yablo has shown that non-circular sets of sentences cause paradox too: Let A i = all sentences ‘A j ’ for j > i are not true. Then {A 0, A 1, A 2,…} are inconsistent.

Yablo’s Paradox Set Y 1 : For all k > 1, Y k is not true. Y 2 : For all k > 2, Y k is not true. Y 3 : For all k > 3, Y k is not true. Y 4 : For all k > 4, Y k is not true. Y 5 : For all k > 5, Y k is not true. … Y n : For all k > n, Y k is not true. …

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…}

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…} All of those guys are false!

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…} All of those guys are false!

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…} All of those guys are false!

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…} All of those guys are false!

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…} All of those guys are false!

Yablo’s Paradox Now consider some number j. Is Y j true or not true? Suppose Y j is true:

Assume Y j is True 1. Y j is trueAssumption 2. For all k > j, Y k is not true.Def of Y j

Yablo’s Paradox Set {… Aj-1, Aj, A j+1, A j+2, A j+3, A j+4, A j+5, A j+6 …} All of those guys are false!

Yablo’s Paradox Set {… Aj-1, Aj, A j+1, A j+2, A j+3, A j+4, A j+5, A j+6 …} All of those guys are false! This particular guy must be false then.

Yablo’s Paradox Set {… Aj-1, Aj, A j+1, A j+2, A j+3, A j+4, A j+5, A j+6 …} All of those guys are false! So what he says must be false.

Yablo’s Paradox Set {… Aj-1, Aj, A j+1, A j+2, A j+3, A j+4, A j+5, A j+6 …} All of those guys are false! So one of these guys must be true.

Yablo’s Paradox Set {… Aj-1, Aj, A j+1, A j+2, A j+3, A j+4, A j+5, A j+6 …} All of those guys are false! So Aj is false too!

Assume Y j is True 1. Y j is trueAssumption 2. For all k > j, Y k is not true.Def of Y j 3. Y j+1 is not true.2 ‘all’ Rule 4. It’s not true that [for all k > j+1, Y k is not true] 3, Def of Y j+1 5. There is some k > j+1 where Y k is true. (2 and 5 are in contradiction)

Thus Y j Are All False The previous argument doesn’t assume anything about Y j. So it works for any number j. Therefore assuming any Y j is true leads to a contradiction. Therefore, all Y j are not true.

Yablo’s Paradox Set {… A j-1, A j, A j+1, A j+2, A j+3, A j+4, A j+5, A j+6 …} All of those guys are false! j could be ANY number

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…} So all of these are false. (They lead to contradictions.)

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…} Thus all of these are false.

Yablo’s Paradox Set {A 0, A 1, A 2, A 3, A 4, A 5, A 6, A 7, A 8,…A j, A j+1,…} All of those guys are false! So what A 0 says is true! (And also, of course, false.)

Thus Y j Are All False But if all Y j are not true, then all Y j for j > 0 are not true. Hence Y 0 is true. But Y 0 is not true, by the previous argument.

5. Accept Some Contradictions In paraconsistent logic, some contradictions are true. Paraconsistent logic denies the (classical) explosion principle, that a contradiction entails anything: Explosion: B & not-B; therefore C Paraconsistent logic claims some sentences (like ‘L’) are both true and false.

Paraconsistent Logic According to paraconsistent logic, there are three (rather than two) possible truth-value assignments to any sentence P.

Three Possibilities TrueFalse P is only T P is T and F P is only F

The “Only a Liar” Sentence But let ‘O’ be defined as follows: O = ‘O’ is false and not true That is, ‘O’ says of itself that it is not one of the sentences that is true and false. It is only false and not also true.

Possibility #1 TrueFalse O is only T O is T and F O is only F

Possibility #1 1. ‘O’ is true and not falseAssumption 2. ‘O’ is true1, ‘and’ Rule 3. O2, Disquotation 4. ‘O’ is false and not true3, Def of O If we say it’s possibility #1, then we have to say it’s possibility #3.

Possibility #2 TrueFalse O is only T O is T and F O is only F

Possibility #2 1. ‘O’ is true and falseAssumption 2. ‘O’ is true1, ‘and’ Rule 3. O2, Disquotation 4. ‘O’ is false and not true3, Def of O If we say it’s possibility #2, then we have to say it’s possibility #3

Possibility #3 TrueFalse O is only T O is T and F O is only F

Possibility #3 1. ‘O’ is false and not trueAssumption 2. O1, Def O 3. ‘O’ is true2, Disquotation 4. ‘O’ is false1, ‘and’ Rule 5. ‘O’ is true and false3,4 ‘and’ Rule If we say it’s #3, it’s #2!

The Liar’s Lesson? There are lots of very complicated solutions to the liar, all of which do one of two things: abandon classical logic or abandon disquotation. It’s clear we have to do one of these things, but neither is very satisfying, and there are no solutions to the liar that everyone likes.

Grelling’s Paradox Grelling’s Paradox or the paradox of heterological terms is very similar to the liar. To begin with, let’s consider a principle like Disquotation, which I’ll just call D2: ‘F’ applies to x = x is F

Examples ‘Dog’ applies to x = x is a dog. ‘Table’ applies to x = x is a table. ‘Philosopher’ applies to x = x is a philosopher. ‘Wednesday’ applies to x = x is a Wednesday. Etc.

Autological and Heterological The analogue of ‘L’ in Grelling’s paradox is the new term ‘heterological’ defined as follows: x is heterological = x does not apply to x We can also define autological, as follows: x is autological = x does apply to x

Examples ‘Short’ applies to ‘short’ ‘English’ applies to ‘English’ ‘Adjectival’ applies to ‘adjectival’ ‘Polysyllabic’ applies to ‘polysyllabic’ So all of these are autological terms.

More Examples ‘Long’ does not apply to ‘long’ ‘German’ does not apply to ‘German’ ‘Nominal’ does not apply to ‘nominal’ ‘Monosyllabic’ does not apply to ‘monosyllabic’ All of these are heterological terms.

Question: Does ‘heterological’ apply to ‘heterological’?

Yes? 1. ‘H’ applies to ‘H’Assumption 2. ‘H’ is H1 D2 3. ‘H’ does not apply to ‘H’2 Def H

No? 1. ‘H’ does not apply to ‘H’Assumption 2. ‘H’ is H1 Def H 3. ‘H’ applies to ‘H’2 D2

Contradiction Just like the liar, we’re led into a contradiction if we assume: D2: ‘F’ applies to x = x is F Law of excluded middle: ‘heterological’ either does or does not apply to itself. A or B, if A then C, if B then C; Therefore, C

Sets There are dogs and cats and couches and mountains and countries and planets. According to Set Theory there are also sets. The set of dogs includes all the dogs as members, and all the members of the set of dogs are dogs. Likewise for the set of mountains, and the set of planets.

Notation To name the set of mountains we write: {x: x is a mountain} “The set of all x such that x is a mountain.” We might introduce a name for this set: M = {x: x is a mountain}

Membership The fundamental relation in set theory is membership, or “being in.” Members of a set are in the set, and non-members are not. Mt. Everest is in {x: x is a mountain}, Michael Jordan is not in {x: x is a mountain}.

Set Theoretic Rules Reduction: a is in {x: COND(x)} Therefore, COND(a) Abstraction: COND(a) Therefore, a is in {x: COND(x)}

Examples Reduction: Mt. Everest is in {x: x is a mountain} Therefore, Mt. Everest is a mountain. Abstraction: Mt. Everest is a mountain. Therefore, Mt. Everest is in {x: x is a mountain}

Self-Membered Sets It’s possible that some sets are members of themselves. Let S = {x: x is a set}. Since S is a set, S is in {x: x is a set} (by abstraction), and thus S is in S (by Def of S). Or consider H = {x: Michael hates x}. Maybe I even hate the set of things I hate. So H is in H.

Russell’s Paradox Set Most sets are non-self-membered. The set of mountains is not a mountain; the set of planets is not a planet; and so on. Define: R = {x: x is not in x}

Is R in R? 1. R is in RYes? 2. R is in {x: x is not in x}1, Def of R 3. R is not in R2, Reduction 4. R is not in RNo? 5. R is in {x: x is not in x}4, Abstraction 6. R is in R5, Def of R

Comparison with the Liar Russell thought that his paradox was of a kind with the liar, and that any solution to one should be a solution to the other. Basically, he saw both as arising from a sort of vicious circularity. If this is right the semantic paradoxes may not be properly “semantic” at all, but arise from a structural feature that many non-semantic paradoxes also have.

Tracking Assumptions To understand Curry’s Paradox, we need to introduce a new notation. In a proof I might wirte: 57. P[Justification] This means that I have proven what’s on line 7, assuming what’s on line 5.

Example 11. L is trueAssumption 12. L 1, Disquotation 13. L is not true2, Def of L Here’s a proof I already did, rewritten. The only assumption I make is in line #1, and what I prove in the other lines assumes what’s on line #1.

Conditional Proof The reason we keep track of assumptions is because some logical rules let us get rid of them. In particular Conditional Proof says that if I assume P and then prove Q, I can conclude [if P then Q] depending on everything Q depends on, except P.

Example 11. L is trueAssumption 12. L 1, Disquotation 13. L is not true2, Def of L 4. If L is true, L is not true1,3 CP In our earlier proof, I could have used CP to show that If L is true, L is not true, resting on no assumptions at all.

Curry’s Paradox Define the Curry sentence C as follows: C = If C is true, then Michael is God.

Curry’s Paradox 11. C is trueAssumption 12. C1, Disquotation 13. If C is true, Michael is God 2, Def of C 14. Michael is God1,3 ‘if’ Rule 5. If C is true, Michael is God 1,4 CP

Curry’s Paradox 5. If C is true, Michael is God 6. C5, Def of C 7. C is true6, Disquotation 8. Michael is God5,7 ‘if’ Rule

SUMMARY