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fix forgotten proof
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CohenCyril committed Aug 8, 2018
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Expand Up @@ -793,6 +793,7 @@ Lemma setU_eq0 A B : (A :|: B == set0) = (A == set0) && (B == set0).
Proof. by rewrite -!subset0 subUset. Qed.

Lemma setD_eq0 A B : (A :\: B == set0) = (A \subset B).
Proof. by rewrite -subset0 subDset setU0. Qed.

Lemma subsetD1 A B x : (A \subset B :\ x) = (A \subset B) && (x \notin A).
Proof.
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