What does fun keyword do in Coq?

Question

I am struggling to understand the meaning of keyword 'fun' in Coq.

There are types all and function forallb:

Inductive all (X : Type) (P : X -> Prop) : list X -> Prop :=
  | all_nil : all X P []
  | all_cons : forall (x:X) (l: list X) , P x -> all X P l -> all X P (x::l).

Fixpoint forallb {X : Type} (test : X -> bool) (l : list X) : bool :=
 match l with
| [] => true
| x :: l' => andb (test x) (forallb test l')
end.

And theorem:

Theorem all_spec: forall (X:Type) (test : X -> bool) (l: list X),
    forallb test l = true <-> all X (fun x => test x = true) l.

I understand the left part but confused on what fun is doing on the right side of the <->.

Answer 1

Isn't it simply like a lambda , ie, isn't here fun x => ... simply like \\x -> ... in Haskell?

There is another interesting peculiarity of that fun ... in your code. The type of the result of that function in your code must be proposition ( Prop ), not boolean . The expression test x = true must be of that type, so we conclude that = in coq denotes propositions about equality, not the boolean binary operation (which is known as == in Haskell; we don't see this from your example, but perhaps coq's notation is similar).

So, although the idea of this fun ... is just a lambda , it is a bit unusual from Haskell's point of view, because here it introduces a function operating on type-level (the result type is Prop ), not value-level (only the latter must be possible--or at least the usual usage--for \\ x-> ... in Haskell). coq's Prop is on the same level as * in Haskell, isn't it?

And all XP in this code is like a type constructor (well, a parameterized family of type constructors) in Haskell, but a dependent one, of type [X] -> * (in Haskell's notation). all_nil and all_cons are like data constructors for this new type.