Theorem dmeqd

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

	Ref	Expression
Hypothesis	dmeqd.1	|- ( ph -> A = B )
Assertion	dmeqd	|- ( ph -> dom A = dom B )

Proof

Step	Hyp	Ref	Expression
1		dmeqd.1	|- ( ph -> A = B )
2		dmeq	|- ( A = B -> dom A = dom B )
3	1 2	syl	|- ( ph -> dom A = dom B )