Metamath Proof Explorer


Theorem bj-rabeqd

Description: Deduction form of rabeq . Note that contrary to rabeq it has no disjoint variable condition. (Contributed by BJ, 27-Apr-2019)

Ref Expression
Hypotheses bj-rabeqd.nf
|- F/ x ph
bj-rabeqd.1
|- ( ph -> A = B )
Assertion bj-rabeqd
|- ( ph -> { x e. A | ps } = { x e. B | ps } )

Proof

Step Hyp Ref Expression
1 bj-rabeqd.nf
 |-  F/ x ph
2 bj-rabeqd.1
 |-  ( ph -> A = B )
3 eleq2
 |-  ( A = B -> ( x e. A <-> x e. B ) )
4 3 anbi1d
 |-  ( A = B -> ( ( x e. A /\ ps ) <-> ( x e. B /\ ps ) ) )
5 2 4 syl
 |-  ( ph -> ( ( x e. A /\ ps ) <-> ( x e. B /\ ps ) ) )
6 1 5 bj-rabbida2
 |-  ( ph -> { x e. A | ps } = { x e. B | ps } )