proposition: Lemma umcntPt p k v : um_count p (pts k v) = if C k && p (k, v) then 1 else 0.
proof: Proof. rewrite /um_count umfiltPt; case C': (C k); last by rewrite dom_undef. by case: ifP; [rewrite domPtK C' | rewrite dom0]. Qed.
imports: Coq(ssreflect ssrbool ssrfun) mathcomp(ssrnat eqtype seq path bigop) pcm(options prelude finmap seqperm pred seqext) pcm(pcm morphism)