proposition: Lemma abs_opp : forall x, abs (opp x) = abs x.
proof: Proof. intros x. apply Rle_antisym. - rewrite opp_mult_m1. rewrite -(Rmult_1_l (abs x)) -abs_opp_one. apply abs_mult. - rewrite -{1}[x]opp_opp. rewrite opp_mult_m1. rewrite -(Rmult_1_l (abs (opp x))) -abs_opp_one. apply abs_mult. Qed.
imports: Coq(Reals Psatz ssreflect) Rcomplements Rbar Markov Iter Lub