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 ( 𝜑𝐹 Fn 𝐴 )
Assertion fnfund ( 𝜑 → Fun 𝐹 )

Proof

Step Hyp Ref Expression
1 fnfund.1 ( 𝜑𝐹 Fn 𝐴 )
2 fnfun ( 𝐹 Fn 𝐴 → Fun 𝐹 )
3 1 2 syl ( 𝜑 → Fun 𝐹 )