Metamath Proof Explorer


Theorem disjeq12i

Description: Equality theorem for disjoint collection. Inference version. (Contributed by GG, 1-Sep-2025)

Ref Expression
Hypotheses disjeq12i.1
|- A = B
disjeq12i.2
|- C = D
Assertion disjeq12i
|- ( Disj_ x e. A C <-> Disj_ x e. B D )

Proof

Step Hyp Ref Expression
1 disjeq12i.1
 |-  A = B
2 disjeq12i.2
 |-  C = D
3 disjeq2
 |-  ( A. x e. A C = D -> ( Disj_ x e. A C <-> Disj_ x e. A D ) )
4 2 a1i
 |-  ( x e. A -> C = D )
5 3 4 mprg
 |-  ( Disj_ x e. A C <-> Disj_ x e. A D )
6 1 disjeq1i
 |-  ( Disj_ x e. A D <-> Disj_ x e. B D )
7 5 6 bitri
 |-  ( Disj_ x e. A C <-> Disj_ x e. B D )