Curry's paradox

In logic, specifically mathematical logic, Curry's paradoxes are a family of logical paradoxes that occur in naive set theory or naive logics. They are named after the logician Haskell Curry.

Related Topics:
Logic - Mathematical logic - Paradox - Naive set theory - Haskell Curry

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An informal version runs as follows:

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:Abelard: "If I'm not mistaken, then Santa Claus exists."

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:Eloise: "I agree: if you are not mistaken then Santa Claus exists."

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:Abelard: "You agree: what I said was correct."

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:Eloise: "Yes."

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:Abelard: "Then I am not mistaken."

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:Eloise: "True."

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:Abelard: "If I am not mistaken, then Santa Claus exists. I am not mistaken. Therefore, Santa Claus exists."

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By this means, any proposition, whether true or not, may be proved.

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The paradox relies on having "Abelard is not mistaken" mean that he is not mistaken about the truth of the entirety of his first statement, including the conditional. Because of the ambiguity of natural language, Eloise could understand something else, but this fact is a red herring and the paradox arises as soon as one allows Abelard to refer to a statement within that very statement. This fact is easier to see once we leave the ambiguity of natural language and try to encode the conversation above using mathematical notation.

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Let us denote by Y the proposition Abelard wants to prove, in this case "Santa Claus exists". Then, let X denote his first statement, which asserts that Y follows from the truth of this first statement. Mathematically, this can be written as X = (X → Y), and we see that X is defined in terms of itself. Is X true?

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If X was true, then X would imply Y. But then, since X is already assumed to be true, Y must also be true. By assuming X, we have just shown Y; in other words, X → Y). Since that statement is equivalent to X, then X is true, and Eloise would have to accept the truth of Abelard's first statement even if she correctly understood what a strange statement he meant.

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Another way of understanding this paradox is to assume that Y is false. Then X becomes X = (X → false), or equivalently (X = not-X). In other words, if Y is false, the proposition X morphs into "this statement is false", which is a contradiction.

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Curry's paradox is one of a group of paradoxical sentences (which also includes the liar paradox) which can be formulated in any language meeting certain conditions. These include: (1) The language must contain apparatus which lets it refer to, and talk about, its own sentences (such as quotation marks, names, or expressions like "this sentence"), and (2) The language must contain its own truth-predicate: that is, the language, call it "L", must contain a predicate meaning "true-in-L", and the ability to ascribe this predicate to any sentences. (Various other sets of conditions are also possible.) Natural languages nearly always contain all these features.

Related Topics:
Liar paradox - Truth

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Logicians are undecided whether such sentences are somehow impermissible (and if so, how to banish them), or meaningless, or whether they are correct and reveal problems with the concept of truth itself (and if so, whether we should reject the concept of truth, or change it), or whether they can be rendered benign by a suitable account of their meanings.

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Note that unlike Russell's paradox, this paradox does not depend on what model of negation is used, as it is completely negation-free. Thus paraconsistent logics still need to take care.

Related Topics:
Russell's paradox - Model of negation - Paraconsistent logics

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The resolution of Curry's paradox is a contentious issue because nontrivial resolutions (such as disallowing X directly) are difficult and not intuitive.

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In set theories which allow unrestricted comprehension, we can prove any logical statement Y from the set

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X equiv left{ x | x in x o Y ight}.

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The proof proceeds:

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egin{matrix}

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mbox{1.} & X in X iff ( X in X o Y ) & mbox{definition of X} \

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mbox{2.} & X in X o ( X in X o Y ) & mbox{from 1} \

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mbox{3.} & X in X o Y & mbox{from 2, contraction} \

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mbox{4.} & (X in X o Y) o X in X & mbox{from 1} \

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mbox{5.} & X in X & mbox{from 3 and 4} \

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mbox{6.} & Y & mbox{from 3 and 5}

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end{matrix}

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