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 V [˙A / x]˙ φ x x = A φ

Proof

Step Hyp Ref Expression
1 df-sbc [˙A / x]˙ φ A x | φ
2 elab6g A V A x | φ x x = A φ
3 1 2 bitrid A V [˙A / x]˙ φ x x = A φ