Metamath Proof Explorer


Theorem rabeqiOLD

Description: Obsolete version of rabeqi as of 3-Jun-2024. (Contributed by Glauco Siliprandi, 26-Jun-2021) Avoid ax-10 and ax-11 . (Revised by Gino Giotto, 20-Aug-2023) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypothesis rabeqi.1
|- A = B
Assertion rabeqiOLD
|- { x e. A | ph } = { x e. B | ph }

Proof

Step Hyp Ref Expression
1 rabeqi.1
 |-  A = B
2 1 nfth
 |-  F/ x A = B
3 eleq2
 |-  ( A = B -> ( x e. A <-> x e. B ) )
4 3 anbi1d
 |-  ( A = B -> ( ( x e. A /\ ph ) <-> ( x e. B /\ ph ) ) )
5 2 4 abbid
 |-  ( A = B -> { x | ( x e. A /\ ph ) } = { x | ( x e. B /\ ph ) } )
6 df-rab
 |-  { x e. A | ph } = { x | ( x e. A /\ ph ) }
7 df-rab
 |-  { x e. B | ph } = { x | ( x e. B /\ ph ) }
8 5 6 7 3eqtr4g
 |-  ( A = B -> { x e. A | ph } = { x e. B | ph } )
9 1 8 ax-mp
 |-  { x e. A | ph } = { x e. B | ph }