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Metamath Proof Explorer

Theorem feqmptd

Description: Deduction form of dffn5 . (Contributed by Mario Carneiro, 8-Jan-2015)

		Ref	Expression
	Hypothesis	feqmptd.1	$⊢ φ \to F : A ⟶ B$
	Assertion	feqmptd	$⊢ φ \to F = (x \in A ⟼ F (x))$

Step	Hyp	Ref	Expression
1		feqmptd.1	$⊢ φ \to F : A ⟶ B$
2	1	ffnd	$⊢ φ \to F Fn A$
3		dffn5	$⊢ F Fn A \leftrightarrow F = (x \in A ⟼ F (x))$
4	2 3	sylib	$⊢ φ \to F = (x \in A ⟼ F (x))$