Metamath Proof Explorer


Theorem dmeqd

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

Ref Expression
Hypothesis dmeqd.1 φ A = B
Assertion dmeqd φ dom A = dom B

Proof

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