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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 ) )