Metamath Proof Explorer


Theorem sbc6g

Description: An equivalence for class substitution. (Contributed by NM, 11-Oct-2004) (Proof shortened by Andrew Salmon, 8-Jun-2011) (Proof shortened by SN, 5-Oct-2024)

Ref Expression
Assertion sbc6g
|- ( A e. V -> ( [. A / x ]. ph <-> A. x ( x = A -> ph ) ) )

Proof

Step Hyp Ref Expression
1 df-sbc
 |-  ( [. A / x ]. ph <-> A e. { x | ph } )
2 elab6g
 |-  ( A e. V -> ( A e. { x | ph } <-> A. x ( x = A -> ph ) ) )
3 1 2 bitrid
 |-  ( A e. V -> ( [. A / x ]. ph <-> A. x ( x = A -> ph ) ) )