Is $\Gamma \vdash x x : T$ possible?

This problem appears on page 104 of Benjamin Pierce's "Types and Programming Languages".

My conclusion is that it is was the case then we would get $x: T_1 \to T_2$ and $x: T_1$ and by some axiom, these types are not equal.

The problem is identifying this axiom but I fear it might be possible to have this equality...

Any hints?

  • $\begingroup$ There's no way to unify $T_1 \to T_2$ and $T_1$ which means (by the typing rule for variables) we would have $x:\tau \in \Gamma \wedge x:\sigma \in \Gamma$ with $\tau \neq \sigma$, which would mean $x$ is not well-typed. $\endgroup$
    – siracusa
    Oct 1, 2019 at 20:46

1 Answer 1


You are on the right track. The argument you would use is on the lines of size of types defined below: (I am assuming you are in the world of simply typed $\lambda$-calculus)

$size(T) = 1$

$size(T \to T') = size(T) + size(T')$

Unification will only work if the size of types is equal, and in this case $size(T \to T') > size(T)$ hence there cannot be such a term.

However, if you add recursion/non-termination in the type system. You can indeed have such a term. $ \vdash (\lambda x. x x) (\lambda x. x x):\bot$

$\bot$ is the type that represents non-termination.


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