Semantic Paradoxes. THE BARBER The Barber Paradox Once upon a time there was a village, and in this...

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 Assumption2. B did not shave any villager X where X shaved X

Assumption3. B did not shave B 1,2 Logic

Suppose B Did Not Shave B

1. B did not shave B Assumption2. B shaved every villager X where X did not shave X

Assumption3. 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.

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 BA implies CB implies C

Therefore, C

Example

Either Michael is dead or he has no legsIf 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.

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 Assumption2. L 1, Disquotation3. ‘L’ is not true 2, Def of L

1 & 3 form a contradiction

“‘L’ is not true”

1. ‘L’ is not true Assumption2. L 1, Def of L3. ‘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:

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 middle2. Give up disjunction elimination3. Give up disquotation4. Disallow self-reference5. 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?

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 trueB: ‘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 true Assumption2. A 1, Disquotation3. ‘B’ is true 2, Def of A4. B 3, Disquotation5. ‘A’ is not true 4, Def of B

“‘A’ is not true”

1. ‘A’ is not true Assumption2. B 1, Def of B3. ‘B’ is true 2, Disquotation4. A 3, Def of A5. ‘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.

Yablo’s Paradox Set

Y1: For all k > 1, Yk is not true.

…Yn: For all k > n, Yk is not true.

{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}

All of those guys are false!

{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}

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

{… Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6…}

This particular guy must be false then.

So what he says must be false.

So one of these guys must be true.

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. 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.

j could be ANY number

{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}

So all of these are false. (They lead to contradictions.)

{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}

Thus all of these 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 FalseP is only T

P is T and FP 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

True FalseO is only T

O is T and FO is only F

Possibility #1

1. ‘O’ is true and not false Assumption2. ‘O’ is true 1, ‘and’ Rule3. O 2, Disquotation4. ‘O’ is false and not true 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 Assumption2. ‘O’ is true 1, ‘and’ Rule3. O 2, Disquotation4. ‘O’ is false and not true 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 Assumption2. O 1, Def O3. ‘O’ is true 2, Disquotation4. ‘O’ is false 1, ‘and’ Rule5. ‘O’ is true and false 3,4 ‘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

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’?

1. ‘H’ applies to ‘H’ Assumption2. ‘H’ is H 1 D23. ‘H’ does not apply to ‘H’ 2 Def H

1. ‘H’ does not apply to ‘H’ Assumption2. ‘H’ is H 1 Def H3. ‘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

RUSSELL’S PARADOX

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 R Yes?2. R is in {x: x is not in x} 1, Def of R3. 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, Abstraction6. 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.

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 Assumption1 2. L 1, Disquotation1 3. L is not true 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 Assumption1 2. L 1, Disquotation1 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.

Curry’s Paradox

Define the Curry sentence C as follows:

C = If C is true, then Michael is God.

Curry’s Paradox

1 1. C is true Assumption1 2. C 1, Disquotation1 3. If C is true, Michael is God

2, Def of C1 4. Michael is God 1,3 ‘if’ Rule

5. If C is true, Michael is God1,4 CP

Curry’s Paradox

5. If C is true, Michael is God6. C 5, Def of C7. C is true 6, Disquotation8. Michael is God 5,7 ‘if’ Rule

SUMMARY

Today we looked at several paradoxes:

• The Barber Paradox• The Liar Paradox• Grelling’s Paradox• Russell’s Paradox• Curry’s Paradox

Semantic Paradoxes. THE BARBER The Barber Paradox Once upon a time there was a village, and in this...

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