Metamath Proof Explorer


Theorem elrabrd

Description: Deduction version of elrab , just like elrabd , but backwards direction. (Contributed by Thierry Arnoux, 15-Jan-2026)

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

Proof

Step Hyp Ref Expression
1 elrabrd.1
 |-  ( x = A -> ( ps <-> ch ) )
2 elrabrd.2
 |-  ( ph -> A e. { x e. B | ps } )
3 1 elrab
 |-  ( A e. { x e. B | ps } <-> ( A e. B /\ ch ) )
4 2 3 sylib
 |-  ( ph -> ( A e. B /\ ch ) )
5 4 simprd
 |-  ( ph -> ch )