proposition: Lemma pval_psub1 x : valid (psub x) -> pval (psub x) = x.
proof: Proof. by rewrite valid_psub=>/andP [H1 H2]; apply: pval_psub. Qed.
imports: Coq(ssreflect ssrbool ssrfun) mathcomp(eqtype fintype) pcm(options axioms prelude) pcm(pcm)