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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
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
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
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
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 true
B: ‘A’ is not true
“‘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
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
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
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
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
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
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
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 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.
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