Metamath Proof Explorer


Theorem funeqi

Description: Equality inference for the function predicate. (Contributed by Jonathan Ben-Naim, 3-Jun-2011)

Ref Expression
Hypothesis funeqi.1 𝐴 = 𝐵
Assertion funeqi ( Fun 𝐴 ↔ Fun 𝐵 )

Proof

Step Hyp Ref Expression
1 funeqi.1 𝐴 = 𝐵
2 funeq ( 𝐴 = 𝐵 → ( Fun 𝐴 ↔ Fun 𝐵 ) )
3 1 2 ax-mp ( Fun 𝐴 ↔ Fun 𝐵 )