Metamath Proof Explorer


Theorem disjeq1d

Description: Equality theorem for disjoint collection. (Contributed by Mario Carneiro, 14-Nov-2016)

Ref Expression
Hypothesis disjeq1d.1
|- ( ph -> A = B )
Assertion disjeq1d
|- ( ph -> ( Disj_ x e. A C <-> Disj_ x e. B C ) )

Proof

Step Hyp Ref Expression
1 disjeq1d.1
 |-  ( ph -> A = B )
2 disjeq1
 |-  ( A = B -> ( Disj_ x e. A C <-> Disj_ x e. B C ) )
3 1 2 syl
 |-  ( ph -> ( Disj_ x e. A C <-> Disj_ x e. B C ) )