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Metamath Proof Explorer

Theorem rabss

Description: Restricted class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006)

Ref Expression
Assertion rabss 
|- ( { x e. A | ph } C_ B <-> A. x e. A ( ph -> x e. B ) )

Proof

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