Metamath Proof Explorer


Theorem disjeq1i

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

Ref Expression
Hypothesis disjeq1i.1
|- A = B
Assertion disjeq1i
|- ( Disj_ x e. A C <-> Disj_ x e. B C )

Proof

Step Hyp Ref Expression
1 disjeq1i.1
 |-  A = B
2 1 rmoeqi
 |-  ( E* x e. A t e. C <-> E* x e. B t e. C )
3 2 albii
 |-  ( A. t E* x e. A t e. C <-> A. t E* x e. B t e. C )
4 df-disj
 |-  ( Disj_ x e. A C <-> A. t E* x e. A t e. C )
5 df-disj
 |-  ( Disj_ x e. B C <-> A. t E* x e. B t e. C )
6 3 4 5 3bitr4i
 |-  ( Disj_ x e. A C <-> Disj_ x e. B C )