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Metamath Proof Explorer

Theorem ss2rab

Description: Restricted abstraction classes in a subclass relationship. (Contributed by NM, 30-May-1999)

Ref Expression
Assertion ss2rab 
|- ( { x e. A | ph } C_ { x e. A | ps } <-> A. x e. A ( ph -> ps ) )

Proof

Step Hyp Ref Expression
1 df-rab 
 |-  { x e. A | ph } = { x | ( x e. A /\ ph ) }
2 df-rab 
 |-  { x e. A | ps } = { x | ( x e. A /\ ps ) }
3 1 2 sseq12i 
 |-  ( { x e. A | ph } C_ { x e. A | ps } <-> { x | ( x e. A /\ ph ) } C_ { x | ( x e. A /\ ps ) } )
4 ss2ab 
 |-  ( { x | ( x e. A /\ ph ) } C_ { x | ( x e. A /\ ps ) } <-> A. x ( ( x e. A /\ ph ) -> ( x e. A /\ ps ) ) )
5 df-ral 
 |-  ( A. x e. A ( ph -> ps ) <-> A. x ( x e. A -> ( ph -> ps ) ) )
6 imdistan 
 |-  ( ( x e. A -> ( ph -> ps ) ) <-> ( ( x e. A /\ ph ) -> ( x e. A /\ ps ) ) )
7 6 albii 
 |-  ( A. x ( x e. A -> ( ph -> ps ) ) <-> A. x ( ( x e. A /\ ph ) -> ( x e. A /\ ps ) ) )
8 5 7 bitr2i 
 |-  ( A. x ( ( x e. A /\ ph ) -> ( x e. A /\ ps ) ) <-> A. x e. A ( ph -> ps ) )
9 3 4 8 3bitri 
 |-  ( { x e. A | ph } C_ { x e. A | ps } <-> A. x e. A ( ph -> ps ) )