proposition: Lemma omfF f k h : f (free h k) = free (f h) k.
proof: Proof. by case: f=>s [m X] /=; rewrite !X omapF. Qed.
imports: Coq(ssreflect ssrbool ssrfun) mathcomp(ssrnat eqtype seq path bigop) pcm(options prelude finmap seqperm pred seqext) pcm(pcm morphism)