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 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

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

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 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:

A or BA implies CB 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 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 trueB: ‘A’ is not true

“‘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

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.

Y2: 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 Y1 is true. But Y1 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 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

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

No?

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

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 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

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

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

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

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.

• $0.10HKD 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 truth-value: 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 →tttttttttttttttttttttttuuuuuuuuuuufffffffffffffffffff

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