What is the proof that the calculus of constructions, extended with recursive types (i.e., Fix at the type-level) isn't strongly normalizing?


With general recursive types you can define the type

type T = T -> T

With that type you can type self-application -- and in fact, every term of the untyped lambda calculus, including any of the well-known fixed-point operators. For example, the Y operator:

Y = \f:T. (\x:T. f (x x)) (\x:T. f (x x))
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    $\begingroup$ It even suffices to have type $T = T \to A$ for some other type $A$. Then $Y I_A = \Omega$ will have type $A$. Of course, this is all because of the negative occurrence of $T$ in type $T \to A$. $\endgroup$ – Andrew Polonsky Apr 16 '17 at 18:01

Here is a slightly more explicit version of what Andreas said.

The term $(\lambda x . x x) (\lambda x . x x)$ is not normalizing because it has exactly one $\beta$-redex, and when we reduce it we get back to the same term. But this term has the type $T$ for any type $T$ satisfying $T = T \to T$.


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