Metamath Proof Explorer

Theorem fnfund

Description: A function with domain is a function, deduction form. (Contributed by Jonathan Ben-Naim, 3-Jun-2011)

Ref Expression
Hypothesis fnfund.1 
|- ( ph -> F Fn A )
Assertion fnfund 
|- ( ph -> Fun F )

Proof

Step Hyp Ref Expression
1 fnfund.1 
 |-  ( ph -> F Fn A )
2 fnfun 
 |-  ( F Fn A -> Fun F )
3 1 2 syl 
 |-  ( ph -> Fun F )