Metamath Proof Explorer


Theorem axfrege52c

Description: Justification for ax-frege52c . (Contributed by RP, 24-Dec-2019)

Ref Expression
Assertion axfrege52c A=B[˙A/x]˙φ[˙B/x]˙φ

Proof

Step Hyp Ref Expression
1 dfsbcq A=B[˙A/x]˙φ[˙B/x]˙φ
2 1 biimpd A=B[˙A/x]˙φ[˙B/x]˙φ