Metamath Proof Explorer


Theorem sbc8g

Description: This is the closest we can get to df-sbc if we start from dfsbcq (see its comments) and dfsbcq2 . (Contributed by NM, 18-Nov-2008) (Proof shortened by Andrew Salmon, 29-Jun-2011) (Proof modification is discouraged.)

Ref Expression
Assertion sbc8g
|- ( A e. V -> ( [. A / x ]. ph <-> A e. { x | ph } ) )

Proof

Step Hyp Ref Expression
1 dfsbcq
 |-  ( y = A -> ( [. y / x ]. ph <-> [. A / x ]. ph ) )
2 eleq1
 |-  ( y = A -> ( y e. { x | ph } <-> A e. { x | ph } ) )
3 df-clab
 |-  ( y e. { x | ph } <-> [ y / x ] ph )
4 equid
 |-  y = y
5 dfsbcq2
 |-  ( y = y -> ( [ y / x ] ph <-> [. y / x ]. ph ) )
6 4 5 ax-mp
 |-  ( [ y / x ] ph <-> [. y / x ]. ph )
7 3 6 bitr2i
 |-  ( [. y / x ]. ph <-> y e. { x | ph } )
8 1 2 7 vtoclbg
 |-  ( A e. V -> ( [. A / x ]. ph <-> A e. { x | ph } ) )