Metamath Proof Explorer


Theorem funin

Description: The intersection with a function is a function. Exercise 14(a) of Enderton p. 53. (Contributed by NM, 19-Mar-2004) (Proof shortened by Andrew Salmon, 17-Sep-2011)

Ref Expression
Assertion funin Fun F Fun F G

Proof

Step Hyp Ref Expression
1 inss1 F G F
2 funss F G F Fun F Fun F G
3 1 2 ax-mp Fun F Fun F G