Metamath Proof Explorer


Theorem rabssf

Description: Restricted class abstraction in a subclass relationship. (Contributed by Glauco Siliprandi, 26-Jun-2021)

Ref Expression
Hypothesis rabssf.1
|- F/_ x B
Assertion rabssf
|- ( { x e. A | ph } C_ B <-> A. x e. A ( ph -> x e. B ) )

Proof

Step Hyp Ref Expression
1 rabssf.1
 |-  F/_ x B
2 df-rab
 |-  { x e. A | ph } = { x | ( x e. A /\ ph ) }
3 2 sseq1i
 |-  ( { x e. A | ph } C_ B <-> { x | ( x e. A /\ ph ) } C_ B )
4 1 abssf
 |-  ( { x | ( x e. A /\ ph ) } C_ B <-> A. x ( ( x e. A /\ ph ) -> x e. B ) )
5 impexp
 |-  ( ( ( x e. A /\ ph ) -> x e. B ) <-> ( x e. A -> ( ph -> x e. B ) ) )
6 5 albii
 |-  ( A. x ( ( x e. A /\ ph ) -> x e. B ) <-> A. x ( x e. A -> ( ph -> x e. B ) ) )
7 df-ral
 |-  ( A. x e. A ( ph -> x e. B ) <-> A. x ( x e. A -> ( ph -> x e. B ) ) )
8 6 7 bitr4i
 |-  ( A. x ( ( x e. A /\ ph ) -> x e. B ) <-> A. x e. A ( ph -> x e. B ) )
9 3 4 8 3bitri
 |-  ( { x e. A | ph } C_ B <-> A. x e. A ( ph -> x e. B ) )