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The Liar Paradox of antiquity goes something like this:

A man says, "Everything I say is a lie."

(I find the modern variation -- "This statement is false" -- to be less interesting. It seems to me to be nothing more than a simple self-contradiction.)

Define 3 logical predicates:

$S(x)$ means $x$ is a sentence

$T(x)$ means $x$ is true

$M(x)$ means the man says $x$

EDITED:

  1. $S(x)\land M(x)\land \forall y(S(y)\rightarrow (M(y)\rightarrow \neg T(y)))$ (Premise)

  2. $S(x)$ (Splitting premise, 1)

  3. $M(x)$

  4. $\forall y(S(y)\rightarrow (M(y)\rightarrow \neg T(y)))$

  5. $S(x)\rightarrow (M(x)\rightarrow\neg T(x))$ (Universal Specification, 4)

  6. $M(x)\rightarrow\neg T(x)$ (Detachment, 2, 5)

  7. $\neg T(x)$ (Detachment, 3, 6)

  8. $\forall a (S(a)\land M(a)\land \forall y(S(y)\rightarrow (M(y)\rightarrow \neg T(y)))\rightarrow \neg T(a))$ (Conclusion, 1)

My question is, does this proof resolve the Liar Paradox?

FOLLOW-UP:

By definition, everything a constant liar says is false. A contradiction arises only, it would seem, when he says something like, "Everything I say is a lie," that is, when he claims:

$$\forall x (S(x)\rightarrow (M(x) \rightarrow \neg T(x))$$

If, as required, this is false, then

$$\exists x (S(x) \land M(x) \land T(x))$$

This contradicts the requirement that everything he says is false. If the constant liar can refrain from making such an admission, no paradoxical situation should arise. No such contradiction would arise, for example, from his saying, "Everything I say is true."

EDITED:

To answer my own question then, the above theorem does not resolve the original Liar Paradox of antiquity. It doesn't "prove" much at all, I'm afraid, but I now feel I have indeed resolved the paradox: It arises from the liar himself claiming that everything he says is false, and my (not necessarily well-founded) assumption that everything he says is indeed false. In hindsight, I think the "prize" must go to Alex Becker for his insightful comment. See my formal proof (Corollary starting on line 19) at http://www.dcproof.com/LiarParadox.htm

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3  
The old liar paradox has never made sense to me. Clearly the man is lying now, but does not always lie. –  Alex Becker Aug 21 '12 at 20:54
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(1) Your version reduces to This statement is false if the man makes only that one statement. (2) What makes This statement is false interesting is that unlike a simple self-contradictory $\varphi\land\lnot\varphi$, it cannot be assigned any truth value. –  Brian M. Scott Aug 21 '12 at 20:57
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There are three sentences in this comment. Exactly two of the sentences are false. You owe me one million dollars. –  Robert Israel Aug 21 '12 at 20:59
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@Dan: Not if it’s the only statement that he ever makes. Then it is precisely equivalent to This statement is a lie. –  Brian M. Scott Aug 21 '12 at 21:15
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@JBeardz: would you recommend any particular reference for that? –  Niel de Beaudrap Aug 22 '12 at 11:37

3 Answers 3

Jack has been a generally unreliable character. He has uttered lots of falsehoods. His regretful last words are "All my assertions (over my whole life) are false".

Is he right? We obviously have to look at what else he has said.

Scenario 1: he has up to his last moments in fact said lots of true things among all the false things. That's enough to make Jack's last words quite unproblematically plain false.

Scenario 2: Jack has, up to his last moments, indeed said nothing but falsehoods. Now trouble. If his final assertion is true it must be because his final assertion is after all yet another falsehood. [Which is as far as the stated argument in the question takes us: but of course the argument now continues ...] However, if Jack's final assertion is false, then indeed all his assertions are false, which is what his last words say, so his last words are true after all. So, in sum, his final words are true if and only if they are false. Contradiction.

Moral: a statement "Everything I say (including this) is false" is not intrinsically paradoxical. But asserted in a suitably unfavourable context it quite certainly can be (i.e. isn't plain false).

For a really good introductory discussion of serious approaches to the Liar paradox, can I recommend http://plato.stanford.edu/entries/liar-paradox/

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If everything you say is false, then you saying so is indeed intrinsically paradoxical. –  Dan Christensen Aug 22 '12 at 4:06
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@DanChristensen You misunderstand. First, the point I was stressing is that "everything I say is false" is plain false in some circumstances, in some circumstances paradoxical. That's why I said the statement is not intrinsically paradoxical: it's status varies in context. –  Peter Smith Aug 22 '12 at 6:28
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@DanChristensen Second, more importantly, what we've shown is that, if "everything I say is false" is indeed one of the things you say, then it can't be that everything you is false (at least if we assume nothing is true and false at the same time). –  Peter Smith Aug 22 '12 at 6:33

It seems to me that your proof is of the Theorem "Any claim by a person M which implies that everything M says is false, is false." That is, at least that one statement is false. That is, M must sometimes speak the truth if they make such a claim. (The claim itself is a lie, but it implies the existence of true utterances.)

This is a straightforward resolution of the paradox, as the sentence involved in the Liar paradox is a universal quantification (as you point out), which implies its own negation, which is an existential quantification over statements.

In effect, as a problem involving speech acts, it represents a case where sarcasm can be decided with no further axioms about the universe of discourse-of-discourse.

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There is no problem with us saying every sentence the constant liar says is false. Logically, there is no problem with him never speaking the truth. As far as I can tell, a contradiction arises only when the constant liar admits that everything he says is false. –  Dan Christensen Aug 22 '12 at 3:32
    
Indeed, as would be the case if they claimed that. "The Liar" here is short for "The Liar Paradox" (which I'm using to refer to "The proposition which is involved in 'The Liar Paradox'"). That is a bit of a severe abbreviation for this level of discussion now that you point it out, though, so I'll make an edit to clarify; I only meant to describe the solution you presented, in English as opposed to logical symbols. –  Niel de Beaudrap Aug 22 '12 at 8:25

(maybe this should be a comment, but it's too long)

(I find the modern variation -- "This statement is false" -- to be less interesting. It seems to me to be nothing more than a simple self-contradiction.)

The modern variation has come into being at least partly because it is a source of serious difficulty in defining formal semantics as well as inspiration for a various number of techniques.

The standard trick is as follows: we call a unary predicate "normal" if it doesn't satisfy itself. This condition is itself a predicate:

Normal(P) := "P(P) is false"

Then we define the statement

S := Normal(Normal)

By the definition of Normal, the literal meaning of S is "Normal(Normal) is false" -- i.e. "S is false". S becomes a liar's statement, and we can go through the usual argument to derive a contradiction:

  • Assume S.
    • Therefore not S
    • Contradiction!
  • Therefore not S.
  • Therefore S
  • Contradiction!

There were no hypotheses to this argument: it proves contradiction is tautology, and so this logic is useless.

The idea of this argument is useful elsewhere. For example, the usual proof the halting problem is uncomputable is essentially the same one as the one given above.

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Sorry, I am not familiar with your notation. –  Dan Christensen Aug 22 '12 at 4:03
    
@DanChristensen: I believe you can read "A := B" as "A is defined to be B". So, e.g., in English, "monosyllabic" or "______ is monosyllabic" are normal predicates (depending on your idea of an English predicate) because "'Monosyllabic' is monosyllabic" is false. –  Rachel Aug 22 '12 at 22:46
    
@Rachel, I am familiar with the ':=' notation. As for the rest, it seems like nonsense. P(P)??? What is that? Can't whatever he is trying to say be expressed in more standard notation? I suspect not. –  Dan Christensen Aug 23 '12 at 4:05
    
@Dan: It is standard notation: formally, predicates are often written as if functions that take objects and return truth values (sometimes, that's how predicates are defined). If you wanted to phrase it in natural language, Rachel has it right. You could define the normal as being an adjective describing those adjectives P that do not have the property P. Sesquipedalian is a good example of an adjective that is not normal. Then, our liar's statement is "normal is normal". –  Hurkyl Aug 23 '12 at 11:07
    
@Hurkyl, "P(P)" is not standard notation. In first-order logic, it is pure nonsense. If you want to be understood here, I suggest you stick to more conventional notation. –  Dan Christensen Aug 23 '12 at 14:30

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