Transcript Semantic Paradoxes - Michael Johnson's Homepage

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
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
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 B
A implies C
B implies C
Therefore, C
Example
Either Michael is dead or he has no legs
If 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.
A or B
A implies C
B implies C
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
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
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 B
A implies C
B implies C
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.
2.
3.
4.
5.
Give up excluded middle
Give up disjunction elimination
Give up disquotation
Disallow self-reference
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?
A or B
A implies C
B implies C
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 true
B: ‘A’ is not true
Circular Reference
‘B’ is
true.
A
B
‘A’ is
false.
Assume ‘A’ Is True
‘B’ is
true.
A
B
‘A’ is
false.
Then ‘B’ Is Also True
‘B’ is
true.
A
B
‘A’ is
false.
But Then ‘A’ is False!
‘B’ is
true.
A
B
‘A’ is
false.
Assume ‘A’ Is False
‘B’ is
true.
A
B
‘A’ is
false.
Then ‘B’ Is Also False
‘B’ is
true.
A
B
‘A’ is
false.
But Then ‘A’ Is Also True
‘B’ is
true.
A
B
‘A’ is
false.
“‘A’ is true”
1. ‘A’ is true
2. A
3. ‘B’ is true
4. B
5. ‘A’ is not true
Assumption
1, Disquotation
2, Def of A
3, Disquotation
4, Def of B
“‘A’ is not true”
1. ‘A’ is not true
2. B
3. ‘B’ is true
4. A
5. ‘A’ is true
Assumption
1, Def of B
2, Disquotation
3, Def of A
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.
Y2: For all k > 2, Yk is not true.
Y3: For all k > 3, Yk is not true.
Y4: For all k > 4, Yk is not true.
Y5: For all k > 5, Yk is not true.
…
Yn: For all k > n, Yk is not true.
…
Yablo’s Paradox Set
{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}
Yablo’s Paradox Set
All of those guys
are false!
{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}
Yablo’s Paradox Set
All of those guys
are false!
{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}
Yablo’s Paradox Set
All of those guys
are false!
{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}
Yablo’s Paradox Set
All of those guys
are false!
{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}
Yablo’s Paradox Set
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
2. For all k > j, Yk is not true.
Assumption
Def of Yj
Yablo’s Paradox Set
All of those guys
are false!
{… Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6…}
Yablo’s Paradox Set
All of those guys
are false!
{… Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6…}
This particular guy
must be false then.
Yablo’s Paradox Set
All of those guys
are false!
{… Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6…}
So what he says must
be false.
Yablo’s Paradox Set
All of those guys
are false!
{… Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6…}
So one of these guys
must be true.
Yablo’s Paradox Set
All of those guys
are false!
{… Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6…}
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.
Yablo’s Paradox Set
All of those guys
are false!
{… Aj-1, Aj, Aj+1, Aj+2, Aj+3, Aj+4, Aj+5, Aj+6…}
j could be ANY number
Yablo’s Paradox Set
{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}
So all of these are false.
(They lead to contradictions.)
Yablo’s Paradox Set
{A0, A1, A2, A3, A4, A5, A6, A7, A8,…Aj, Aj+1,…}
Thus all of these are false.
Yablo’s Paradox Set
All of those guys
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
P is only T
P is T and F
P is only F
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
True
O is only T
O is T and F
O is only F
False
Possibility #1
1. ‘O’ is true and not false
2. ‘O’ is true
3. O
4. ‘O’ is false and not true
Assumption
1, ‘and’ Rule
2, Disquotation
3, Def of O
If we say it’s possibility #1, then we have to say
it’s possibility #3.
Possibility #2
True
O is only T
O is T and F
O is only F
False
Possibility #2
1. ‘O’ is true and false
2. ‘O’ is true
3. O
4. ‘O’ is false and not true
Assumption
1, ‘and’ Rule
2, Disquotation
3, Def of O
If we say it’s possibility #2, then we have to say
it’s possibility #3
Possibility #3
True
O is only T
O is T and F
O is only F
False
Possibility #3
1. ‘O’ is false and not true
2. O
3. ‘O’ is true
4. ‘O’ is false
5. ‘O’ is true and false
If we say it’s #3, it’s #2!
Assumption
1, Def O
2, Disquotation
1, ‘and’ Rule
3,4 ‘and’ Rule
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’?
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
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
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
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
1
1. L is true
2. L
3. L is not true
Assumption
1, Disquotation
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
1
1. L is true
2. L
3. L is not true
4. If L is true, L is not true
Assumption
1, Disquotation
2, Def of L
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
1
1
1. C is true
Assumption
2. C
1, Disquotation
3. If C is true, Michael is God
2, Def of C
4. Michael is God
1,3 ‘if’ Rule
5. If C is true, Michael is God
1,4 CP
Curry’s Paradox
5. If C is true, Michael is God
6. C
5, Def of C
7. C is true
6, Disquotation
8. 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