Semantic Paradoxes THE BARBER The Barber Paradox Once

Semantic Paradoxes

THE BARBER

The Barber Paradox Once upon a time there was a village, and in this village lived a barber named B. B shaved all the villagers who did not shave themselves, And B shaved none of the villagers who did shave themselves. Question, did B shave B, or not?

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

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

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

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 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.

THE PARADOX OF THE LIAR

Epimenides was a Cretan who famously said “All Cretans lie all the time. ” If what he said was true, then it was a lie, and thus false. If what he said was false, then it was not a lie, and thus true. His statement is false if and only if it is true.

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. ” We’ll stick with: L = ‘L’ is not true

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

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

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

Solutions • • Give up excluded middle Give up disquotation Disallow self-reference Accept that some contradictions are true

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 true at the same time. ”

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

“‘A’ is true” 1. ‘A’ is true 2. A 3. ‘B’ is true 4. B 5. ‘A’ is not true Assumption 1, Disquotation 2, Def of A 3, Disquotation 4, Def of B

“‘A’ is not true” 1. ‘A’ is not true 2. B 3. ‘B’ is true 4. A 5. ‘A’ is true Assumption 1, Def of B 2, Disquotation 3, Def of A 4, Disquotation

Disallowing Circular Reference Even circular reference is not essential. Stephen Yablo has shown that non-circular sets of sentences cause paradox too: Let Ai = all sentences ‘Aj’ 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, Yk is not true. Y 2: For all k > 2, Yk is not true. … Yn: For all k > n, Yk is not true. …

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

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

Thus Yj Are All False The previous argument doesn’t assume anything about Yj. So it works for any number j. Therefore assuming any Yj is true leads to a contradiction. Therefore, all Yj are not true. But if all Yj are not true, then all Yj for j > 1 are not true. Hence Y 1 is true. But Y 1 is not true, by the previous argument.

Paraconsistent Logic 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.

Three Possibilities According to paraconsistent logic, there are three (rather than two) possible truth-value assignments to any sentence P: • P is true and not false (P is only true) • P is true and false • P is false and not true (P is only false)

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

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

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

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 D 2: ‘F’ applies to x = x is F

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’ 2. ‘H’ is H 3. ‘H’ does not apply to ‘H’ Assumption 1 D 2 2 Def H

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

Contradiction Just like the liar, we’re led into a contradiction if we assume: • D 2: ‘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

RUSSELL’S PARADOX

Sets There are dogs and cats and couches and mountains and countries and planets. According to Set Theory there 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 R 2. R is in {x: x is not in x} 3. R is not in R Yes? 1, Def of R 2, Reduction 4. R is not in R 5. R is in {x: x is not in x} 6. R is in R No? 4, Abstraction 5, 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.

CURRY’S PARADOX

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

Example 1 1. L is true 2. L 3. L is not true Assumption 1, Disquotation 2, 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 1 1. L is true 2. L 3. L is not true 4. If L is true, L is not true Assumption 1, Disquotation 2, Def of L 1, 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 1 1 1. C is true Assumption 2. C 1, Disquotation 3. If C is true, Michael is God 2, Def of C 4. Michael is God 1, 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. C 5, Def of C 7. C is true 6, Disquotation 8. Michael is God 5, 7 ‘if’ Rule

THE PARADOX OF THE HEAP

Sorites 1. 1 grain of sand is not a heap. 2. For all numbers n: if n grains of sand are not a heap, then n + 1 grains of sand are not a heap. 3. Therefore, 200 trillion grains of sand are not a heap.

The Other Way 1. 200 trillion grains of sand makes a heap. 2. For all numbers n: if n + 1 grains of sand make a heap, then n grains of sand make a heap. 3. Therefore 1 grain of sand makes a heap.

What To Do? Neither of these sorites arguments results in a contradiction… until you add in the obvious fact that the conclusion of each is false. To deny the conclusion, we need to deny either premise 1 or premise 2 or logic.

Denying Premise 1 In the first argument, premise 1 is: 1 grain of sand is not a heap. In the second it’s: 200 trillion grains of sand is a heap.

Denying Premise 2 (Argument 1) says: For all numbers n: if n grains of sand are not a heap, then n + 1 grains of sand are not a heap. The negation of this is: There exists a number n such that: n grains of sand are not a heap, but n + 1 grains of sand are a heap.

Denying Premise 2 (Argument 2) says: For all numbers n: if n + 1 grains of sand make a heap, then n grains of sand make a heap. The negation of this is: There is a number n such that: n + 1 grains of sand make a heap, but n grains of sand do not make a heap.

No Sharp Boundaries Premise 2 in both cases asserts No Sharp Boundaries. It’s never true that one grain of sand makes the difference between a heap and not a heap.

No Sharp Boundaries • One hair doesn’t make the difference between being bald and not bald. • One micrometer doesn’t make the difference between being tall and not tall. • $0. 10 HKD does not make the difference between being rich and not rich. • One nanosecond does not make the difference between being old and not old.

Epistemicism One solution is to claim that there ARE sharp boundaries, but we can never know where they are. Acquiring $0. 10 can make someone go from not rich to rich, but we can’t ever know when this happens. Basic problem: What determines the boundary if not how we use the words? What determines how we use the words if not what we (can) know?

Many-Valued Logics Another solution is to introduce a new truthvalue: True, False, and Undefined. There’s No Sharp Boundaries, because there’s no point at which adding one hair moves someone from truly bald to falsely bald. More hairs → ttttttttttttuuuuuuffffffffff

Higher-Order Vagueness The problem is that now there are sharp boundaries between being truly bald and undefinedly bald, and between being undefinedly bald, and falsely bald. Intuitively, adding one hair to a truly bald person can’t make them undefinedly bald.

SUMMARY

Today we looked at several paradoxes: • The Barber Paradox • The Liar Paradox • Grelling’s Paradox • Russell’s Paradox • Curry’s Paradox • The Paradox of the Heap