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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 )