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 )