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 ) )