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