# Semantic Paradoxes - Michael Johnson

Semantic Paradoxes 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 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 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. Its the Law

Either its Tuesday or its not Tuesday. Either its Wednesday or its 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 cant run the race. If Michael has no legs, he cant run the race.

Therefore, Michael cant 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.

THE PARADOX OF THE LIAR 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 true

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

1. L is not true Assumption 2. L 1, Def of L 3. L is true 2, 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. 2. 3. 4. 5.

Give up excluded middle Give up disjunction elimination Give up disquotation Disallow self-reference 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 isnt essential to the paradox. A: B is true B: A is not true Circular Reference B is true.

A B A is false.

Assume A Is True B is true. A B

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

A B A is false.

But Then A is False! B is true. A B

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

A B A is false.

Then B Is Also False B is true. A B

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

A B A is false.

A is true 1. A is true Assumption 2. A 1, Disquotation 3. B is true 2, Def of A 4. B

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

1, Def of B 3. B is true 2, Disquotation 4. A 3, Def of A 5. A is true 4, 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 Ai = all sentences Aj for j > i are not true. Then {A0, A1, A2,} are inconsistent. Yablos Paradox Set Y1: For all k > 1, Yk is not true.

Y2: For all k > 2, Yk is not true. Y3: For all k > 3, Yk is not true. Y4: For all k > 4, Yk is not true. Y5: For all k > 5, Yk is not true. Yn: For all k > n, Yk is not true.

Yablos Paradox Set {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,} Yablos Paradox Set All of those guys are false! {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,}

Yablos Paradox Set All of those guys are false! {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,} Yablos Paradox Set All of those guys

are false! {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,} Yablos Paradox Set All of those guys are false! {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,}

Yablos Paradox Set All of those guys are false! {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,} Yablos 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

Yablos Paradox Set All of those guys are false! { Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6} Yablos Paradox Set All of those guys

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

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

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

are false! { Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6} So Aj is false too! 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. Its 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 doesnt 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. Yablos Paradox Set All of those guys are false! { Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6}

j could be ANY number Yablos Paradox Set {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,} So all of these are false. (They lead to contradictions.)

Yablos Paradox Set {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,} Thus all of these are false. Yablos Paradox Set

All of those guys are false! {A0, A1, A2, A3, A4, A5, A6, A7, A8,Aj, Aj+1,} So what A0 says is true! (And also, of course, false.) Thus Yj Are All False

But if all Yj are not true, then all Yj for j > 0 are not true. Hence Y0 is true. But Y0 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 True P is only T

P is T and F P is only F 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 True O is only T

O is T and F O is only F False Possibility #1 1. O is true and not false Assumption

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

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

False Possibility #2 1. O is true and false Assumption 2. O is true 1, and Rule 3. O

2, Disquotation 4. O is false and not true 3, Def of O If we say its possibility #2, then we have to say its possibility #3 Possibility #3

True O is only T O is T and F O is only F False Possibility #3

1. O is false and not true Assumption 2. O 1, Def O 3. O is true 2, Disquotation 4. O is false 1, and Rule

5. O is true and false 3,4 and Rule If we say its #3, its #2! The Liars 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.

Its 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. GRELLINGS PARADOX Grellings Paradox Grellings Paradox or the paradox of

heterological terms is very similar to the liar. To begin with, lets consider a principle like Disquotation, which Ill 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 Grellings 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 H 1 D2 3. H does not apply to H

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

2 D2 Contradiction Just like the liar, were 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 RUSSELLS PARADOX 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 Its 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. Russells 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 Yes? 2. R is in {x: x is not in x} 1, Def of R 3. R is not in R 2, Reduction

4. R is not in R No? 5. R is in {x: x is not in x} 4, Abstraction 6. R is in R 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.

CURRYS PARADOX Tracking Assumptions To understand Currys Paradox, we need to introduce a new notation. In a proof I might wirte: 5 7. P

[Justification] This means that I have proven whats on line 7, assuming whats on line 5. Example 1 1. L is true

Assumption 1 2. L 1, Disquotation 1 3. L is not true 2, Def of L Heres a proof I already did, rewritten. The only assumption I make is in line #1, and what I prove in the other lines assumes whats 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 Assumption 1 2. L 1, Disquotation 1 3. L is not true 2, Def of L

4. If L is true, L is not true 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. Currys Paradox Define the Curry sentence C as follows: C = If C is true, then Michael is God.

Currys Paradox 1 1. C is true Assumption 1 2. C 1, Disquotation 1 3. If C is true, Michael is God 2, Def of C

1 4. Michael is God 1,3 if Rule 5. If C is true, Michael is God 1,4 CP Currys 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 SUMMARY