semantic paradoxes continued

Semantic Paradoxes Continued

RECAP

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

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.

Disjunction Elimination

A or BA implies CB implies C

Therefore, C

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.

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.

Disquotation

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

P = ‘P’ is true

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

‘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

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

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

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

We write out sets by putting names of their members between brackets. So the set of full professors in Lingnan philosophy is:

{Darell, Neven, Paisley}

Notation

We can also write the set using a condition:

{x: x is a full professor in Lingnan philosophy}

This is the same as the set {Darell, Neven, Paisley}. We might introduce a name for this set:

F = {x: x is a full professor in Lingnan philosophy}

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

Historical Importance

Russell’s paradox was what caused Frege to stop doing mathematics and do philosophy of language instead.

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.

Comparison with the Liar

If this is right the semantic paradoxes may not be properly “semantic” at all, but arise from a structural feature that many non-semantic things (like sets) also have.

The von Neumann Heirarchy

CURRY’S PARADOX

Haskell Brooks Curry

• Mathematician who worked on combinatory logic.

• Has three computer languages named after him: Haskell, Brooks, and Curry.

• Devised a semantic paradox.

Conditional Proof

Suppose you want to prove a conditional (“if-then”) statement.

For example, suppose you want to show that if the accuser is telling the truth, then the accused should go to jail.

The Accusation

Michael kicked me.

Assuming for the Sake of Argument

First, you would assume for the sake of argument that the accuser is telling the truth. Assume that Michael did in fact kick the puppy.

(Even though of course he’s innocent.)

Conditional Proof

Then you would use that assumption to show that Michael belonged in jail.

You would argue that since kicking puppies violates article 2, section 6, paragraph 3 of the criminal code, Michael belongs in jail.

Conditional Proof

Finally, you would stop assuming that Michael did actually kick the puppy and conclude:

If the accuser is telling the truth, then Michael belongs in jail.

Modus Ponens

There’s one other rule of logic that involves conditionals. This rule is used when we already know a conditional is true:

Premise: if A, then BPremise: A

Conclusion: B

Curry’s Paradox

Define the Curry sentence C as follows:

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

Proving C

To prove C we do a conditional proof:

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

Proving C

To prove C we do a conditional proof:

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

Assume this.

Proving C

To prove C we do a conditional proof:

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

Prove this.

Proving C

To prove C we do a conditional proof:

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

Throw out assumption and conclude this.

Proof of C

1. ‘C’ is true Assumption2. C 1, Disquotation3. ‘C’ is true, then Michael is God

2, Definition of C4. Michael is God 1, 3, Modus Ponens

Proof of C

1. ‘C’ is true Assumption2. C 1, Disquotation3. ‘C’ is true, then Michael is God

2, Definition of C4. Michael is God 1, 3, Modus Ponens5. If ‘C’ is true, then Michael is God

1, 4 Cond. Proof

No Paradox Yet

Now we have proven C. This in itself should not bother us.

We are not committed to saying Michael is God, only that if ‘C’ is true, then Michael is God.

Proof that Michael is God

1. If ‘C’ is true, then Michael is God.Previous Proof

2. C 1, Definition of C3. ‘C’ is true 2, Disquotation4. Michael is God 1, 3, Modus Ponens

Set Theory Version

Curry’s Paradox also comes in a set theoretic flavor. The paradoxical set is this one:

M = {x: if x is in x, then Michael is God}

Requirements

Curry’s Paradox leads to contradiction (or absurdity, or anything you like) and all it requires is:• Self-reference• Disquotation• Conditional Proof• Modus Ponens

Important Feature

Curry’s Paradox uses different logical rules (modus ponens, conditional proof) than the liar paradox (excluded middle, disjunction elimination).

This suggests that it’s probably not the logic, but instead self-reference and/ or disquotation that is the problem.

Important Feature

Recall that paraconsistent logics can avoid the Liar Paradox by assuming that some sentences can be both true and false while avoiding explosion.

Even with such an assumption, however, Curry’s Paradox still works, and leads to explosion, since you can prove anything with it.

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.

Paradox

Neither of these sorites arguments results in a contradiction… until you add in the obvious fact that the conclusion of each is false.

Borderline Cases

The paradox seems to arise whenever we have a term that admits of borderline cases.

There are some people that I don’t know whether they’re rich out of uncertainty: because I don’t know how much money they have. These are not borderline cases.

Borderline Cases

The paradox seems to arise whenever we have a term that admits of borderline cases.

But there are other people that I don’t know whether they are rich even though I know exactly how much money they have. These are borderline cases.

Borderline Cases

Most of our ordinary language admits of borderline cases:

Big, tall, short, rich, fast, slow, smart, dumb, funny, long, flat, narrow…

Also: mountain, car, tree, horse…

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.

Solutions

1. Accept Sharp Boundaries.2. Introduce more truth-values.

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.

Epistemicism

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?

Epistemicism

Further problem: the epistemicist says we can’t know where the Sharp Boundary is, but that it exists. However, he has to admit that we can:• Guess where the Sharp Boundary is.• Wonder where the Sharp Boundary is.• Fear that we are crossing the Sharp Boundary

(e.g. for getting old).But all these seem silly!

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.

Many-Valued Logics

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.

Many-Valued Logics

More hairs →tttttttttttttttttttttttuuuuuuuuuuufffffffffffffffffff

Two sharp boundaries!

Fuzzy Logic

Instead, we might try having infinitely many truth-values: 1 is fully true, 0 is fully false, and any number in between is less than fully true.

More hairs →1 1 1 1 1 1 .99 .98 .98 .97… .12 .11 .1 .1 0 0 0 0 0

Fuzzy Logic

A fuzzy logician has to explain how to calculate the truth-values of complex expressions from the truth values of their parts. Common rules:• The truth-value of “~P” is 1 minus the truth-

value of P• The truth-value of “P & Q” is the lowest of the

truth-values of P and Q.• The truth-value of “P or Q” is the highest of the

truth values of P and Q.

Problems

“P & ~P” should always be fully false: 0.

But if P = 0.5, then “P & ~P” = 0.5