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Metamath Proof Explorer

Theorem iineq12dv

Description: Equality deduction for indexed intersection. (Contributed by Glauco Siliprandi, 26-Jun-2021) Remove DV conditions. (Revised by GG, 1-Sep-2025)

Ref Expression
Hypotheses iineq12dv.1 
|- ( ph -> A = B )
iineq12dv.2 
|- ( ( ph /\ x e. B ) -> C = D )
Assertion iineq12dv 
|- ( ph -> |^|_ x e. A C = |^|_ x e. B D )

Proof

Step Hyp Ref Expression
1 iineq12dv.1 
 |-  ( ph -> A = B )
2 iineq12dv.2 
 |-  ( ( ph /\ x e. B ) -> C = D )
3 1 eleq2d 
 |-  ( ph -> ( x e. A <-> x e. B ) )
4 3 imbi1d 
 |-  ( ph -> ( ( x e. A -> t e. C ) <-> ( x e. B -> t e. C ) ) )
5 4 ralbidv2 
 |-  ( ph -> ( A. x e. A t e. C <-> A. x e. B t e. C ) )
6 5 abbidv 
 |-  ( ph -> { t | A. x e. A t e. C } = { t | A. x e. B t e. C } )
7 df-iin 
 |-  |^|_ x e. A C = { t | A. x e. A t e. C }
8 df-iin 
 |-  |^|_ x e. B C = { t | A. x e. B t e. C }
9 6 7 8 3eqtr4g 
 |-  ( ph -> |^|_ x e. A C = |^|_ x e. B C )
10 2 iineq2dv 
 |-  ( ph -> |^|_ x e. B C = |^|_ x e. B D )
11 9 10 eqtrd 
 |-  ( ph -> |^|_ x e. A C = |^|_ x e. B D )