Metamath Proof Explorer


Theorem sbc5

Description: An equivalence for class substitution. (Contributed by NM, 23-Aug-1993) (Revised by Mario Carneiro, 12-Oct-2016) (Proof shortened by SN, 2-Sep-2024)

Ref Expression
Assertion sbc5
|- ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )

Proof

Step Hyp Ref Expression
1 df-sbc
 |-  ( [. A / x ]. ph <-> A e. { x | ph } )
2 clelab
 |-  ( A e. { x | ph } <-> E. x ( x = A /\ ph ) )
3 1 2 bitri
 |-  ( [. A / x ]. ph <-> E. x ( x = A /\ ph ) )