Open:
/Coq/Sets/Powerset_facts/Union_commutative.con

We prove the conjunction again:

alias U /Coq/Sets/Ensembles/Ensembles/U.var
alias V /Coq/Sets/Powerset_facts/Sets_as_an_algebra/U.var
alias Ensemble /Coq/Sets/Ensembles/Ensemble.con
alias Union    /Coq/Sets/Ensembles/Union.ind#1/1
alias Included /Coq/Sets/Ensembles/Included.con
alias and      /Coq/Init/Logic/and.ind#1/1

The two parts of the conjunction can be proved in the same way. So we
can make a Cut:

!C:Ensemble{U:=V}.!D:Ensemble{U:=V}.
 (Included{U:=V} (Union{U:=V} C D) (Union{U:=V} D C))