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Metamath Proof Explorer

Theorem dmeqd

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

		Ref	Expression
	Hypothesis	dmeqd.1	$⊢ φ \to A = B$
	Assertion	dmeqd	$⊢ φ \to dom A = dom B$

Step	Hyp	Ref	Expression
1		dmeqd.1	$⊢ φ \to A = B$
2		dmeq	$⊢ A = B \to dom A = dom B$
3	1 2	syl	$⊢ φ \to dom A = dom B$