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Metamath Proof Explorer

Theorem fnfund

Description: A function with domain is a function, deduction form. (Contributed by Jonathan Ben-Naim, 3-Jun-2011)

		Ref	Expression
	Hypothesis	fnfund.1	$⊢ φ \to F Fn A$
	Assertion	fnfund	$⊢ φ \to Fun F$

Step	Hyp	Ref	Expression
1		fnfund.1	$⊢ φ \to F Fn A$
2		fnfun	$⊢ F Fn A \to Fun F$
3	1 2	syl	$⊢ φ \to Fun F$