How to match a "match" expression? - coq

Question

I am trying to write a rule for hypotheses, formulated with a help of match construction:
Goal forall x:nat, (match x with | 1 => 5 | _ => 10 end = 5 -> x = 1)%nat.
intros.
x : nat
H : match x with
| 0%nat => 10%nat
| 1%nat => 5%nat
| S (S _) => 10%nat
end = 5%nat
============================
x = 1%nat
How can I match such hypotheses ? The following straight forward method fails:
match goal with
|[H:match ?e with | ?a => ?x | ?b => ?y | ?c => ?z end = ?b] => idtac
end.
> Syntax error: 'end' expected after [branches] (in [match_constr]).

Answer 1

Pattern-matching on match statements is somewhat weird.
The first thing you should know is that, inside Coq, there's no such thing as a match on several variables or with deep matching: everything is translated in terms of simpler match statements. Thus, the term you wrote is actually syntax sugar for the following term:
match x with
| 0 => 10
| S x' =>
match x' with
| 0 => 5
| S x'' => 10
end
end
which is what Coq is hinting at when it prints your proof state. The problem is that this syntax sugar doesn't work on Ltac patterns: thus, when writing an Ltac pattern that mentions a match, you should always try to match it as if it were a one-level match.
The second problem is that you can't bind the pattern part of a match: something like
match goal with
| H : match ?x => _ | ?y => _ end = 5 |- _ => (* ... *)
end
doesn't really make sense in Ltac.
You have two choices for solving your problem, then:
Write down the match you expect with the exact list of constructors of your type on the pattern part, e.g.
match goal with
| H : match x with 0 => _ | S _ => _ end = 5 |- _ => (* ... *)
end
Use the special match (* ... *) with _ => _ end syntax, which matches any match whatsoever:
match goal with
| H : match x with _ => _ end = 5 |- _ => (* ... *)
end
Often, as in your case, one still wants to consider all branches of match, including deep ones. This idiom often comes in handy in that case:
repeat match goal with
| H : match ?x with _ => _ end = _ |- _ =>
destruct x; try solve [inversion H]
end.