The Coq mailing list

[Arnaud Spiwack <aspiwack AT lix.polytechnique.fr>]

Well, technically [and_rect] and all other the [*_rect]-s are defined (using pattern-matching). You may want to [Print and_rect] to see what it looks like. You can, of course, define modus ponens like this:

Definition modus_ponens {P Q:Prop} (L:P->Q) (h:P) : Q := L h.

And, more generally, for forall (in Coq, [P->Q] is just a notation for [forall _:P,Q]):

Definition forall_rect {A:Type} {Q:A->Prop} (L:forall y, Q y) (x:A) : Q x := L x.

The fundamental rules of Coq (it's an arbitrary choice if you will, but it's technically motivated) are: "forall" is a type, inductive types are types, values of type "forall" are contructed with "fun" and eliminated with application, values of inductive types are constructed with constructors and eliminated with "match".

On 7 August 2014 07:25, Terrell, Jeffrey <jeffrey.terrell AT kcl.ac.uk> wrote:

In some texts, -> is treated in the same way as any other connective; and given that e.g. /\ has an elimination rule, i.e. and_rect, I was wondering whether -> has an elimination rule too. The same applies to forall x : A, P.

It's clear that there's something fundamental about type P -> Q, because attempting to define it in the normal way is problematic.

Inductive imply (P Q : Prop) : Prop :=

   imp : (P -> Q) -> imply P Q.

Regarding forall x : A, P, I'm aware that you can build a proof f a of P[a/x], but what is it that says you can?

Regards,

Jeff.

On 6 Aug 2014, at 22:09, Maxime Dénès wrote:

Well, given a proof a of A and a proof f of forall x : A, P, you can build the proof f a of P[a/x].

Not sure if this answer your question, though.

Are you familiar with the Curry-Howard isomorphism? If not, I would advise to read about it.

Best,

Maxime.

On 08/06/2014 04:59 PM, Terrell, Jeffrey wrote:

From a proof of A and a proof of A -> B, we can infer B. This is the -> elimination rule.

Similarly, from a proof a of A and a proof of forall x : A, P, we can infer P[a/x]. This is the forall elimination rule.

How are these elimination rules manifest in Coq?

Regards,

Jeff.