Does the Y combinator contradict the Curry-Howard correspondence?

The Y combinator has the type $(a \rightarrow a) \rightarrow a$. By the Curry-Howard Correspondence, because the type $(a \rightarrow a) \rightarrow a$ is inhabited, it must correspond to a true theorem. However $a \rightarrow a$ is always true, so it appears as if the Y combinator's type corresponds to the theorem $a$, which is not always true. How can this be?

The original Curry-Howard correspondence is an isomorphism between intuitionistic propositional logic and the simply-typed lambda calculus.

There are, of course, other Curry-Howard-like isomorphisms; Phil Wadler famously pointed out that the double-barrelled name "Curry-Howard" predicts other double-barrelled names like "Hindley-Milner" and "Girard-Reynolds". It would be funny if "Martin-Löf" was one of them, but it isn't. But I digress.

The Y combinator does not contradict this, for one key reason: it is not expressible in the simply-typed lambda calculus.

In fact, that was the whole point. Haskell Curry discovered the fixpoint combinator in the untyped lambda calculus, and used it to prove that the untyped lambda calculus is not a sound deduction system.

Interestingly, the type of Y corresponds to a logical paradox which isn't as well-known as it should be, called Curry's paradox. Consider this sentence:


If this sentence is true, then Santa Claus exists.

Suppose the sentence were true. Then, clearly, Santa Claus would exist. But this is precisely what the sentence says, so the sentence is true. Therefore, Santa Claus exists. QED

StackExchange Computer Science Q#32497, answer score: 24