Metamath Proof Explorer


Theorem fmpttd

Description: Version of fmptd with inlined definition. Domain and codomain of the mapping operation; deduction form. (Contributed by Glauco Siliprandi, 23-Oct-2021) (Proof shortened by BJ, 16-Aug-2022)

Ref Expression
Hypothesis fmpttd.1
|- ( ( ph /\ x e. A ) -> B e. C )
Assertion fmpttd
|- ( ph -> ( x e. A |-> B ) : A --> C )

Proof

Step Hyp Ref Expression
1 fmpttd.1
 |-  ( ( ph /\ x e. A ) -> B e. C )
2 eqid
 |-  ( x e. A |-> B ) = ( x e. A |-> B )
3 1 2 fmptd
 |-  ( ph -> ( x e. A |-> B ) : A --> C )