How to pull the rhs out of an equality in coq - coq

Question

If I have the following:
H : some complicated expression = some other complicated expression
and I want to grab
u := some other complicated expression
without hardcoding it into my proof (i.e., using pose)
Is there a clean way to do this in LTac?

Answer 1

I am sure there are other ltac ways to do it, in my case I prefer to use SSReflect's contextual pattern language to do it. (You'll need to install the plugin or use Coq >= 8.7 which includes SSReflect):
(* ce_i = complicated expression i *)
Lemma example T (ce_1 ce_2 : T) (H : ce_1 = ce_2) : False.
set u := (X in _ = X) in H.
resulting goal:
T : Type
ce_1, ce_2 : T
u := ce_2 : T
H : ce_1 = u
============================
False
Usually you can refine the pattern more and more until you get a pretty stable match.
Note that this happens to be the first example of the section 8.3 "Contextual patterns" in the SSReflect manual.

Answer 2

Here is another version, which uses Ltac and its ability to pattern-match on types of terms:
Tactic Notation "assign" "rhs" "of" ident(H) "to" ident(u) "in" ident(H') :=
match type of H with _ = ?rhs => set (u := rhs) in H' end.
Tactic Notation "assign" "rhs" "of" ident(H) "to" ident(u) "in" "*" :=
match type of H with _ = ?rhs => set (u := rhs) in * end.
We can create more variants of the above (see e.g. here). Here is how to use it:
Lemma example {T} (ce1 ce2 ce3 : T) (H1 : ce1 = ce2) (H2 : ce2 = ce3) : ce1 = ce3.
Proof.
assign rhs of H1 to u in *.
Proof state:
u := ce2 : T
H1 : ce1 = u
H2 : u = ce3
============================
ce1 = ce3
One more time:
Undo.
assign rhs of H1 to u in H1.
Proof state:
u := ce2 : T
H1 : ce1 = u
H2 : ce2 = ce3
============================
ce1 = ce3