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 AV[˙A/x]˙φxx=Aφ

Proof

Step Hyp Ref Expression
1 df-sbc [˙A/x]˙φAx|φ
2 elab6g AVAx|φxx=Aφ
3 1 2 bitrid AV[˙A/x]˙φxx=Aφ