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Metamath Proof Explorer

Theorem ssab

Description: Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006)

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

Proof

Step Hyp Ref Expression
1 abid2 
 |-  { x | x e. A } = A
2 1 sseq1i 
 |-  ( { x | x e. A } C_ { x | ph } <-> A C_ { x | ph } )
3 ss2ab 
 |-  ( { x | x e. A } C_ { x | ph } <-> A. x ( x e. A -> ph ) )
4 2 3 bitr3i 
 |-  ( A C_ { x | ph } <-> A. x ( x e. A -> ph ) )