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Metamath Proof Explorer

Theorem fnfun

Description: A function with domain is a function. (Contributed by NM, 1-Aug-1994)

Ref Expression
Assertion fnfun ( 𝐹 Fn 𝐴 → Fun 𝐹 )

Proof

Step Hyp Ref Expression
1 df-fn ⊢ ( 𝐹 Fn 𝐴 ↔ ( Fun 𝐹 ∧ dom 𝐹 = 𝐴 ) )
2 1 simplbi ⊢ ( 𝐹 Fn 𝐴 → Fun 𝐹 )