Metamath Proof Explorer


Theorem fexd

Description: If the domain of a mapping is a set, the function is a set. (Contributed by Glauco Siliprandi, 26-Jun-2021)

Ref Expression
Hypotheses fexd.1
|- ( ph -> F : A --> B )
fexd.2
|- ( ph -> A e. C )
Assertion fexd
|- ( ph -> F e. _V )

Proof

Step Hyp Ref Expression
1 fexd.1
 |-  ( ph -> F : A --> B )
2 fexd.2
 |-  ( ph -> A e. C )
3 fex
 |-  ( ( F : A --> B /\ A e. C ) -> F e. _V )
4 1 2 3 syl2anc
 |-  ( ph -> F e. _V )