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.

  • $\begingroup$ To start with, do you understand intuitively why there should be no term with that type? $\endgroup$ Commented Jan 20, 2020 at 16:16
  • $\begingroup$ Kind of, basically it would mean that if such a term existed then the term would belong to all possible types. $\endgroup$
    – Apoorv
    Commented Jan 20, 2020 at 16:25
  • 3
    $\begingroup$ It would be more accurate to say that such a term would represent a function where you give it any type and it gives you back a value of that type. So you could substitute any uninhabited type for $t$ in the term of type $\forall t. t$ to get a value of that uninhabited typeโ€”a contradiction. (This is not a complete proof by itself because you would also need to show that System F has an uninhabited type.) $\endgroup$ Commented Jan 20, 2020 at 16:36

3 Answers 3


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 $\cdot\vdash t : \forall \alpha.\alpha$.

  • $\begingroup$ Thank you for your answer! I can see the structure of the proof now. Can you please help me understand how do you read the equation [[โˆ€๐›ผ.๐ต]]๐›พ :=[[๐ต]](๐—๐—‹๐—Ž๐–พ,๐›พ)โˆง[[๐ต]](๐–ฟ๐–บ๐—…๐—Œ๐–พ,๐›พ) $\endgroup$
    – Apoorv
    Commented Jan 21, 2020 at 15:07
  • $\begingroup$ @ApoorvIngle the line is the definition of the Bool^n -> Bool function in the case when a type is a forall-quantified type. It says that a quantified type yields true in a truth-valuation environment if it yields true no matter how the bound alpha variable is interpreted (i.e. as true or as false). $\endgroup$ Commented Jan 21, 2020 at 15:34
  • 1
    $\begingroup$ We interpret each closed type as either true (a "true proposition") or false (a "false proposition"). Since a universally quantified type ranges over arbitrary types, we have in the model that a forall-type is true iff it's true when it is instantiated with a true proposition plus also true when it is instantiated with a false proposition. $\endgroup$ Commented Jan 21, 2020 at 15:39

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 โˆ€ฮฑ.ฮฑ 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, โˆ€ฮฑ.ฮฑ 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


Here is a simple answer. If we have a lambda-term $x : \forall t.\,t$ then we should be able to apply the term $x$ to any type $t$ and obtain a value of type $t$. Denote type application by $x t$ and find that $x t$ is a value of any given type $t$. Let us apply this ability to the void type $0$. Then $x 0$ is a value of type $0$. But there are no values of type $0$. So, we could not have had a value $x$ either.

Another proof is based on using naturality laws. If you have a pure lambda-term of type $\forall t.\,t$ in System F, you can conclude by parametricity theorems that this term satisfies the law of natural transformations of type $\forall t.\,\underline 1 \to t$ (where I denote the unit type by $\underline1$).

Then you write the naturality law: for an arbitrary function $f: a\to b$, a function $\phi: \forall t.\,\underline1\to t$ satisfies the equation:

$$ f \circ \phi = \phi $$

We can derive from this equation that there can't be any function $\phi$. For an $x: \underline1$ (there is only one such value $x$) we will have some value $a_1=\phi(x)$ of type $a$ and also some value $b_1=\phi(x)$ of type $b$. However, the value $b_1$ must be at the same time equal to $f(a_1)$ for all functions $f: a\to b$. This is impossible. So, we conclude that $\phi$ cannot exist, and the type $\forall t.\,t$ is void.


Your Answer

By clicking โ€œPost Your Answerโ€, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.