Metamath Proof Explorer


Theorem iineq1i

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

Ref Expression
Hypothesis iineq1i.1
|- A = B
Assertion iineq1i
|- |^|_ x e. A C = |^|_ x e. B C

Proof

Step Hyp Ref Expression
1 iineq1i.1
 |-  A = B
2 1 eleq2i
 |-  ( x e. A <-> x e. B )
3 2 imbi1i
 |-  ( ( x e. A -> t e. C ) <-> ( x e. B -> t e. C ) )
4 3 ralbii2
 |-  ( A. x e. A t e. C <-> A. x e. B t e. C )
5 4 abbii
 |-  { t | A. x e. A t e. C } = { t | A. x e. B t e. C }
6 df-iin
 |-  |^|_ x e. A C = { t | A. x e. A t e. C }
7 df-iin
 |-  |^|_ x e. B C = { t | A. x e. B t e. C }
8 5 6 7 3eqtr4i
 |-  |^|_ x e. A C = |^|_ x e. B C