2 /Coq/Sets/Powerset_facts/Union_commutative.con
4 We prove the conjunction again:
6 alias U /Coq/Sets/Ensembles/Ensembles/U.var
7 alias V /Coq/Sets/Powerset_facts/Sets_as_an_algebra/U.var
8 alias Ensemble /Coq/Sets/Ensembles/Ensemble.con
9 alias Union /Coq/Sets/Ensembles/Union.ind#1/1
10 alias Included /Coq/Sets/Ensembles/Included.con
11 alias and /Coq/Init/Logic/and.ind#1/1
13 The two parts of the conjunction can be proved in the same way. So we
16 !C:Ensemble{U:=V}.!D:Ensemble{U:=V}.
17 (Included{U:=V} (Union{U:=V} C D) (Union{U:=V} D C))