Metamath Proof Explorer


Theorem rabeqbii

Description: Equality theorem for restricted class abstractions. Inference version. (Contributed by GG, 1-Sep-2025)

Ref Expression
Hypotheses rabeqbii.1
|- A = B
rabeqbii.2
|- ( ph <-> ps )
Assertion rabeqbii
|- { x e. A | ph } = { x e. B | ps }

Proof

Step Hyp Ref Expression
1 rabeqbii.1
 |-  A = B
2 rabeqbii.2
 |-  ( ph <-> ps )
3 1 eleq2i
 |-  ( x e. A <-> x e. B )
4 3 2 anbi12i
 |-  ( ( x e. A /\ ph ) <-> ( x e. B /\ ps ) )
5 4 abbii
 |-  { x | ( x e. A /\ ph ) } = { x | ( x e. B /\ ps ) }
6 df-rab
 |-  { x e. A | ph } = { x | ( x e. A /\ ph ) }
7 df-rab
 |-  { x e. B | ps } = { x | ( x e. B /\ ps ) }
8 5 6 7 3eqtr4i
 |-  { x e. A | ph } = { x e. B | ps }