proposition: Lemma gcdr_def : forall x a b, x %| a -> x %| b -> (forall x', x' %| a -> x' %| b -> (x' %| x)) -> gcdr a b %= x.
proof: Proof. by move=> x a b xa xb hx; rewrite eqd_def dvdr_gcd xa xb hx. Qed.
imports: HB(structures) Coq(ssreflect ssrfun ssrbool Arith.Wf_nat) mathcomp(eqtype ssrnat div seq path) mathcomp(ssralg fintype perm tuple choice generic_quotient) mathcomp(matrix bigop zmodp mxalgebra poly) stronglydiscrete