Metamath Proof Explorer


Theorem ffdmd

Description: The domain of a function. (Contributed by Glauco Siliprandi, 26-Jun-2021)

Ref Expression
Hypothesis ffdmd.1
|- ( ph -> F : A --> B )
Assertion ffdmd
|- ( ph -> F : dom F --> B )

Proof

Step Hyp Ref Expression
1 ffdmd.1
 |-  ( ph -> F : A --> B )
2 ffdm
 |-  ( F : A --> B -> ( F : dom F --> B /\ dom F C_ A ) )
3 1 2 syl
 |-  ( ph -> ( F : dom F --> B /\ dom F C_ A ) )
4 3 simpld
 |-  ( ph -> F : dom F --> B )