Why is the type ∀t.t un-inhabited in System F?

Question

How do you prove that there exists no term with the type $\forall t. t$ in System F?

I tried searching through Pierce's TAPL and Reynold's ToPL, but could not find anything. I suspect that the proof may involve some kind of model theoritic argument, but I fail to see where to start.

Answer 1

The simplest proof is giving a model where types are interpreted as propositions, and terms as proofs, then observing that $\forall \alpha.\alpha$ is interpreted as the false proposition, so any $\cdot \vdash t : \forall \alpha.\alpha$ would be a proof of falsehood. It is important to remember that $\forall \alpha.\alpha$ is only uninhabited for sure in the empty context $\cdot$, in other contexts it may be trivially inhabited.

Concretely, first interpret each $A$ type with $n$ free type variables in the following way:

$$ \begin{alignat*}{2} & [\![ A ]\!] && : \mathsf{Bool}^n \to \mathsf{Bool}\\ & [\![ \alpha ]\!]\,\gamma && := \text{$\alpha$-th component of $\gamma$}\\ & [\![ A \to B ]\!]\,\gamma && := [\![ A ]\!]\,\gamma \Rightarrow [\![ B ]\!]\,\gamma\\ & [\![ \forall \alpha.\,B]\!]\,\gamma && := [\![B]\!]\,(\mathsf{true},\,\gamma)\,\land\,[\![B]\!]\,(\mathsf{false},\,\gamma) \end{alignat*} $$

Interpret every typing context $\Gamma$ with $n$ free type variables as:

$$ \begin{alignat*}{2} & [\![\Gamma]\!] : \mathsf{Bool}^n\to \mathsf{Bool}\\ & [\![\Gamma]\!]\,\gamma := \bigwedge\limits_{A\,\in\,\Gamma}\,[\![A]\!]\,\gamma \end{alignat*} $$

Then show for each $\Gamma \vdash t : A$ term, that for each truth valuation $\gamma : \mathsf{Bool}^n$, if $[\![\Gamma]\!]\,\gamma = \mathsf{true}$, then $[\![A]\!]\,\gamma = \mathsf{true}$. This can be done by induction on terms.

Now, $[\![\forall \alpha.\alpha]\!]$ with no free type variables evaluates to $\mathsf{false}$, since it's $\mathsf{true}$ when $\alpha$ is instantiated to $\mathsf{true}$, and $\mathsf{false}$ when it's instantiated to $\mathsf{false}$, and the conjunction of these is $\mathsf{false}$. Assuming $\cdot\vdash t : \forall \alpha.\alpha$, we have by the previous result that $[\![\forall \alpha.\alpha]\!]$ is $\mathsf{true}$, a contradiction, hence there is no $\cdot\vdash t : \forall \alpha.\alpha$.

Answer 2

There is a proof by using parametricity/logical relations framework or free theorems as mentioned in Zhao et al.[1]

For instance, we can conclude that there is no closed inhabitant of type ∀α.α in a pure setting. If there were such a term, it must yield a value of any type at which it is instantiated, but there is no uniform algorithm to compute a value at any type. Therefore, ∀α.α is an empty type.

[1]: Zhao J., Zhang Q., Zdancewic S. (2010) Relational Parametricity for a Polymorphic Linear Lambda Calculus. In: Ueda K. (eds) Programming Languages and Systems. APLAS 2010. Lecture Notes in Computer Science, vol 6461. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-642-17164-2_24