Semantic - PowerPoint Presentation

Slide1

Semantic Paradoxes ContinuedSlide2

RecapSlide3

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?Slide4

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 LogicSlide5

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

Slide6

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.Slide7

Disjunction Elimination

A or B

A implies C

B implies C

Therefore, CSlide8

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, contradictionSlide9

Contradictions

Whenever we are confronted with a contradiction, we need to

give up

something that led us into the contradiction. Slide10

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.Slide11

Disquotation

To say P is the same thing as saying ‘P’ is true. This is the “

disquotation

principle”:

P

=

‘P’

is

trueSlide12

Liar Sentence

L

= ‘L’ is not trueSlide13

“‘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 contradictionSlide14

“‘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 contradictionSlide15

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.Slide16

Solutions

Give up excluded middle

Give up disjunction elimination

Give up

disquotation

Disallow self-reference

Accept that some contradictions are trueSlide17

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.Slide18

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 FSlide19

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 xSlide20

Question: Does ‘

heterological

’ apply to ‘

heterological

’?Slide21

Yes?

1. ‘H’ applies to ‘H’ Assumption

2. ‘H’ is H 1 D2

3. ‘H’ does not apply to ‘H’ 2

Def

HSlide22

No?

1. ‘H’ does not apply to ‘H’ Assumption

2. ‘H’ is H 1

Def

H

3. ‘H’ applies to ‘H’ 2 D2Slide23

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, CSlide24

Russell’s paradoxSlide25

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.Slide26

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}Slide27

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

}Slide28

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}.Slide29

Set Theoretic Rules

Reduction:

a is in {x: COND(x)}

Therefore, COND(a)

Abstraction:

COND(a)

Therefore, a is in {x: COND(x)}Slide30

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}Slide31

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.Slide32

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}Slide33

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 RSlide34

Historical Importance

Russell’s paradox was what caused

Frege

to stop doing mathematics and do philosophy of language instead.Slide35

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.Slide36

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.Slide37

The von Neumann HeirarchySlide38

Curry’s paradoxSlide39

Haskell Brooks Curry

Mathematician who worked on combinatory logic.

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

Devised a semantic paradox.Slide40

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.Slide41

The Accusation

Michael kicked me.Slide42

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.)Slide43

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.Slide44

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.Slide45

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 B

Premise: A

Conclusion: BSlide46

Curry’s Paradox

Define the Curry sentence C as follows:

C = If C is true, then Michael is God.Slide47

Proving C

To prove

C we do a conditional proof:

C = If

‘C’

is true, then Michael is God.Slide48

Proving C

To prove

C we do a conditional proof:

C = If

‘

C’

is true

, then Michael is God.

Assume this.Slide49

Proving C

To prove

C we do a conditional proof:

C = If

‘C’

is true, then

Michael is God

.

Prove this.Slide50

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.Slide51

Proof of C

‘C’ is true Assumption

C 1,

Disquotation

‘C’ is true, then Michael is God

2, Definition of C

4. Michael is God 1, 3, Modus PonensSlide52

Proof of C

‘C’ is true Assumption

C 1,

Disquotation

‘C’ is true, then Michael is God

2, Definition of C

4. Michael is God 1, 3, Modus Ponens

5.

If ‘C’ is true, then Michael is God

1, 4 Cond. ProofSlide53

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.Slide54

Proof that Michael is God

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

Previous Proof

2. C 1, Definition of C

3. ‘C’ is true 2,

Disquotation

4. Michael is God 1, 3, Modus PonensSlide55

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}Slide56

Requirements

Curry’s Paradox leads to contradiction (or absurdity, or anything you like) and all it requires is:

Self-reference

Disquotation

Conditional Proof

Modus PonensSlide57

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.Slide58

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.Slide59

the paradox of the heapSlide60

Sorites

1 grain of sand is not a heap.

For all numbers n: if n grains of sand are not a heap, then n + 1 grains of sand are not a heap.

Therefore, 200 trillion grains of sand are not a heap.Slide61

The Other Way

200 trillion grains of sand makes a heap.

For all numbers n: if n + 1 grains of sand make a heap, then n grains of sand make a heap.

Therefore 1 grain of sand makes a heap.Slide62

Paradox

Neither

of these

sorites

arguments results in a contradiction… until you add in the obvious fact that the conclusion of each is false.Slide63

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.Slide64

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.Slide65

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…Slide66

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.Slide67

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.Slide68

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.Slide69

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.Slide70

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.Slide71

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.Slide72

Solutions

Accept Sharp Boundaries.

Introduce more truth-values.Slide73

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. Slide74

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?Slide75

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!Slide76

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. Slide77

Many-Valued Logics

More hairs →

tttttttttttttttttttttttuuuuuuuuuuufffffffffffffffffffSlide78

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.Slide79

Many-Valued Logics

More hairs →

tttttttttttttttttttttttuuuuuuuuuuufffffffffffffffffff

Two sharp boundaries!Slide80

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 0Slide81

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.Slide82

Problems

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

But if P = 0.5, then “P & ~P” = 0.5