Metamath Proof Explorer


Theorem disjeq1f

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

Ref Expression
Hypotheses disjss1f.1
|- F/_ x A
disjss1f.2
|- F/_ x B
Assertion disjeq1f
|- ( A = B -> ( Disj_ x e. A C <-> Disj_ x e. B C ) )

Proof

Step Hyp Ref Expression
1 disjss1f.1
 |-  F/_ x A
2 disjss1f.2
 |-  F/_ x B
3 eqimss2
 |-  ( A = B -> B C_ A )
4 2 1 disjss1f
 |-  ( B C_ A -> ( Disj_ x e. A C -> Disj_ x e. B C ) )
5 3 4 syl
 |-  ( A = B -> ( Disj_ x e. A C -> Disj_ x e. B C ) )
6 eqimss
 |-  ( A = B -> A C_ B )
7 1 2 disjss1f
 |-  ( A C_ B -> ( Disj_ x e. B C -> Disj_ x e. A C ) )
8 6 7 syl
 |-  ( A = B -> ( Disj_ x e. B C -> Disj_ x e. A C ) )
9 5 8 impbid
 |-  ( A = B -> ( Disj_ x e. A C <-> Disj_ x e. B C ) )