Metamath Proof Explorer


Theorem sbc7

Description: An equivalence for class substitution in the spirit of df-clab . Note that x and A don't have to be distinct. (Contributed by NM, 18-Nov-2008) (Revised by Mario Carneiro, 13-Oct-2016)

Ref Expression
Assertion sbc7
|- ( [. A / x ]. ph <-> E. y ( y = A /\ [. y / x ]. ph ) )

Proof

Step Hyp Ref Expression
1 sbccow
 |-  ( [. A / y ]. [. y / x ]. ph <-> [. A / x ]. ph )
2 sbc5
 |-  ( [. A / y ]. [. y / x ]. ph <-> E. y ( y = A /\ [. y / x ]. ph ) )
3 1 2 bitr3i
 |-  ( [. A / x ]. ph <-> E. y ( y = A /\ [. y / x ]. ph ) )