Coq has a type Prop of proof irrelevant propositions which are discarded during extraction. What are the reason for having this if we use Coq only for proofs. Prop is impredicative, so Prop : Prop, however, Coq automatically infers universe indexes and we can use Type(i) instead everywhere. It seems Prop complicates everything a lot.

I read that there're philosophical reasons for separating Set and Prop in Luo's book, however, I didn't find them in the book. What are they?

  • 6
    $\begingroup$ “if we use Coq only for proofs”: I think you've identified a key point here. Coq isn't used only for proofs. $\endgroup$ Commented Mar 31, 2014 at 14:48

4 Answers 4


$\mathtt{Prop}$ is very useful for program extraction because it allows us to delete parts of code that are useless. For example, to extract a sorting algorithm we would prove the statement "for every list $\ell$ there is a list $k$ such that $k$ is ordered and $k$ is a permutatiom of $\ell$". If we write this down in Coq and extract without using $\mathtt{Prop}$, we will get:

  1. "for all $\ell$ there is $k$" will give us a map sort which takes lists to lists,
  2. "such that $k$ is ordered" will give a funciton verify which runs through $k$ and checks that it is sorted, and
  3. "$k$ is a permutation of $\ell$" will give a permutation pi which takes $\ell$ to $k$. Note that pi is not just a mapping, but also the inverse mapping together with programs verifying that the two maps really are inverses.

While the extra stuff is not totally useless, in many applications we want to get rid of it and keep just sort. This can be accomplished if we use $\mathtt{Prop}$ to state "$k$ is ordered" and "$k$ is a permutation of $\ell$", but not "for all $\ell$ there is $k$".

In general, a common way to extract code is to consider a statement of the form $\forall x : A \,.\, \exists y : B \,.\, \phi(x, y)$ where $x$ is input, $y$ is output, and $\phi(x,y)$ explains what it means for $y$ to be a correct output. (In the above example $A$ and $B$ are the types of lists and $\phi(\ell, k)$ is "$k$ is ordered and $k$ is a permutation of $\ell$.") If $\phi$ is in $\mathtt{Prop}$ then extraction gives a map $f : A \to B$ such that $\phi(x, f(x))$ holds for all $x \in A$. If $\phi$ is in $\mathtt{Set}$ then we also get a function $g$ such that $g(x)$ is the proof that $\phi(x, f(x))$ holds, for all $x \in A$. Often the proof is computationally useless and we prefer to get rid of it, especially when it is nested deeply inside some other statement. $\mathtt{Prop}$ gives us the possibility to do so.

Added 2015-07-29: There is a question whether we could avoid $\mathsf{Prop}$ altogether by automatically optimizing away "useless extracted code". To some extent we can do that, for instance all code extracted from the negative fragment of logic (stuff built from the empty type, unit type, products) is useless as it just shuffles around the unit. But there are genuine design decisions one has to make when using $\mathsf{Prop}$. Here is a simpe example, where $\Sigma$ means that we are in $\mathsf{Type}$ and $\exists$ means we are in $\mathsf{Prop}$. If we extract from $$\Pi_{n : \mathbb{N}} \Sigma_{b : \{0,1\}} \Sigma_{k : \mathbb{N}} \; n = 2 \cdot k + b$$ we will get a program which decomposes $n$ into its lowest bit $b$ and the remaining bits $k$, i.e., it computes everything. If we extract from $$\Pi_{n : \mathbb{N}} \Sigma_{b : \{0,1\}} \exists_{k : \mathbb{N}} \; n = 2 \cdot k + b$$ then the program will only compute the lowest bit $b$. The machine cannot tell which is the correct one, the user has to tell it what they want.

  • 2
    $\begingroup$ I'm slightly confused. Are you saying that without $\mathtt{Prop}$ it'd be impossible to recognize in the extracted program that $g(x)$ doesn't contribute to the output (i.e. that it merely verifies it)? Are there scenarios where one wouldn't be able to pull out such useless code through the usual means available to code optimizers? $\endgroup$
    – user
    Commented Jul 24, 2015 at 7:13
  • 1
    $\begingroup$ From the extracted program one could say, "I want $k$," and backtrack from there. I haven't been able to come up with a scenario so entangled that we couldn't optimize away anything that doesn't directly contribute to determining the permutation without it in fact being necessary for computing said permutation (from a global optimization standpoint, anyway). $\endgroup$
    – user
    Commented Jul 29, 2015 at 4:47
  • 3
    $\begingroup$ You do not have the information "I want $k$". That is an extra assumption, and obviously once they tell you which particular result they want, you can just optimize away dead code. Actually, I thought of a better answer: it is a design question which things to put in $\mathsf{Prop}$. You need to know what the user wants, and he tells you what he wants by using $\mathsf{Prop}$. It is easy to come up with examples where there are several options. I will add one to my answer. $\endgroup$ Commented Jul 29, 2015 at 9:46
  • 2
    $\begingroup$ As far as I know nobody can really tell how to extract anything from $(-1)$-types. It's clear that they contain some computational content, but not what this might be. $\endgroup$ Commented Jul 29, 2015 at 14:24
  • 4
    $\begingroup$ Ah, okay. Using $\mathtt{Prop}$ as a way of specifying design decisions makes a lot more sense to me than as a way of deleting useless code. $\endgroup$
    – user
    Commented Jul 29, 2015 at 18:12

$\mathrm{Prop}$ is impredicative, which create a very expressive proof system. However it is "too" expressive in the following sense:

$$ \mathrm{impredicative\ Prop} + \mathrm{large\ elimination} + \mathrm{excluded\ middle} $$

is inconsistent. Usually you want to keep the possibility to add the excluded middle, so one solution is to keep large elimination and make Prop predicative. The other is to suppress large elimination.

Coq did both! They renamed the predicative Prop to Set, and disabled large elimination in Prop.

The expressiveness gained by impredicativity is "reassuring" in the sense 99% of "reasonable" mathematics can be formalized with it, and it is known to be consistent relative to set theory. This makes it likely they won't weaken it to something like Agda, which only has predicative universes.

  • 10
    $\begingroup$ Oh and I forgot to mention: it is not the case that Prop : Prop, that would be inconsistent. Rather quantifications over all propositions is again a proposition. $\endgroup$
    – cody
    Commented Apr 1, 2014 at 21:31
  • $\begingroup$ Could you point me to any more resources about this? Everything I can find seems very scattered. $\endgroup$
    – user833970
    Commented Dec 21, 2017 at 20:17
  • 2
    $\begingroup$ @user833970 any specific things you'd like pointers to? I'm afraid there isn't really an all encompassing reference for meta theory of dependent types. This discussion (which points back here!) might be useful: github.com/FStarLang/FStar/issues/360 $\endgroup$
    – cody
    Commented Dec 23, 2017 at 9:36
  • $\begingroup$ thank, I'm working through the Berardi paradox paper now, I think that will clear up my confusion. $\endgroup$
    – user833970
    Commented Dec 23, 2017 at 22:48

Even if you are not interested in extracting programs, the fact that Prop is impredicative allows you to build some models which you can't build using a predicative tower of universes. IIRC Thorsten Altenkirch has a model of System F using Coq's impredicativity.


The principle of propositions-as-types (or formulas-as-types), also known as the Curry-Howard correspondence, is the key idea for viewing (intuitionistic) type theories as logical systems and to apply type theories (and constructive mathematics in general) to computer science. Suppose U is the type of all propositions, then U is also the type of all types by the identification. Therefore, type U of all types naturally occurs and, in particular, it is the type of itself. In Martin-Löf's (predicative) type theory, propositions and types are still identified, but the type of all types is replaced by an infinite sequence of type universes U0 U 1: U2..... Hence, one can not quantify over all propositions or predicates, although quantification over each universe is allowed.

That's the reason type and prop are separated. Without this separation the resulting system becomes inconsistent.


Your Answer

By clicking “Post Your Answer”, 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.