Metamath Proof Explorer


Theorem funeqd

Description: Equality deduction for the function predicate. (Contributed by NM, 23-Feb-2013)

Ref Expression
Hypothesis funeqd.1 ⊢ φ → A = B
Assertion funeqd ⊢ φ → Fun ⁡ A ↔ Fun ⁡ B

Proof

Step Hyp Ref Expression
1 funeqd.1 ⊢ φ → A = B
2 funeq ⊢ A = B → Fun ⁡ A ↔ Fun ⁡ B
3 1 2 syl ⊢ φ → Fun ⁡ A ↔ Fun ⁡ B