Transcript The 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
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
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 B
A implies C
B 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
2. L
3. ‘L’ is not true
1 & 3 form a contradiction
Assumption
1, Disquotation
2, Def of L
“‘L’ is not true”
1. ‘L’ is not true
2. L
3. ‘L’ is true
1 & 3 form a contradiction
Assumption
1, Def of L
2, Disquotation
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
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’
2. ‘H’ is H
3. ‘H’ does not apply to ‘H’
Assumption
1 D2
2 Def H
No?
1. ‘H’ does not apply to ‘H’
2. ‘H’ is H
3. ‘H’ applies to ‘H’
Assumption
1 Def 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 nonmembers 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
2. R is in {x: x is not in x}
3. R is not in R
Yes?
1, Def of R
2, Reduction
4. R is not in R
5. R is in {x: x is not in x}
6. R is in R
No?
4, Abstraction
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 nonsemantic 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 (“ifthen”) 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 B
Premise: 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
Assumption
2. C
1, Disquotation
3. ‘C’ is true, then Michael is God
2, Definition of C
4. Michael is God
1, 3, Modus Ponens
Proof of C
1. ‘C’ is true
Assumption
2. C
1, Disquotation
3. ‘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. 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 C
3. ‘C’ is true
2, Disquotation
4. 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 truthvalue: 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 truthvalue 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