proposition: Lemma fsfun_with_id f x : [fsfun f with x |-> f x] = f.
proof: Proof. by apply/fsfunP=> z; rewrite fsfun_withE; case: eqP => [->|]. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrnat eqtype ssrfun seq) mathcomp(choice path finset finfun fintype bigop)