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Metamath Proof Explorer

Theorem rneqd

Description: Equality deduction for range. (Contributed by NM, 4-Mar-2004)

		Ref	Expression
	Hypothesis	rneqd.1	$⊢ φ \to A = B$
	Assertion	rneqd	$⊢ φ \to ran A = ran B$

Step	Hyp	Ref	Expression
1		rneqd.1	$⊢ φ \to A = B$
2		rneq	$⊢ A = B \to ran A = ran B$
3	1 2	syl	$⊢ φ \to ran A = ran B$