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