Metamath Proof Explorer

Theorem funeqd

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

Ref Expression
Hypothesis funeqd.1 
|- ( ph -> A = B )
Assertion funeqd 
|- ( ph -> ( Fun A <-> Fun B ) )

Proof

Step Hyp Ref Expression
1 funeqd.1 
 |-  ( ph -> A = B )
2 funeq 
 |-  ( A = B -> ( Fun A <-> Fun B ) )
3 1 2 syl 
 |-  ( ph -> ( Fun A <-> Fun B ) )