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Metamath Proof Explorer

Theorem ssrabf

Description: Subclass of a restricted class abstraction. (Contributed by Glauco Siliprandi, 26-Jun-2021)

Ref Expression
Hypotheses ssrabf.1 
|- F/_ x B
ssrabf.2 
|- F/_ x A
Assertion ssrabf 
|- ( B C_ { x e. A | ph } <-> ( B C_ A /\ A. x e. B ph ) )

Proof

Step Hyp Ref Expression
1 ssrabf.1 
 |-  F/_ x B
2 ssrabf.2 
 |-  F/_ x A
3 df-rab 
 |-  { x e. A | ph } = { x | ( x e. A /\ ph ) }
4 3 sseq2i 
 |-  ( B C_ { x e. A | ph } <-> B C_ { x | ( x e. A /\ ph ) } )
5 1 ssabf 
 |-  ( B C_ { x | ( x e. A /\ ph ) } <-> A. x ( x e. B -> ( x e. A /\ ph ) ) )
6 1 2 dfss3f 
 |-  ( B C_ A <-> A. x e. B x e. A )
7 6 anbi1i 
 |-  ( ( B C_ A /\ A. x e. B ph ) <-> ( A. x e. B x e. A /\ A. x e. B ph ) )
8 r19.26 
 |-  ( A. x e. B ( x e. A /\ ph ) <-> ( A. x e. B x e. A /\ A. x e. B ph ) )
9 df-ral 
 |-  ( A. x e. B ( x e. A /\ ph ) <-> A. x ( x e. B -> ( x e. A /\ ph ) ) )
10 7 8 9 3bitr2ri 
 |-  ( A. x ( x e. B -> ( x e. A /\ ph ) ) <-> ( B C_ A /\ A. x e. B ph ) )
11 4 5 10 3bitri 
 |-  ( B C_ { x e. A | ph } <-> ( B C_ A /\ A. x e. B ph ) )