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