Metamath Proof Explorer


Theorem dmeqd

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

Ref Expression
Hypothesis dmeqd.1 φA=B
Assertion dmeqd φdomA=domB

Proof

Step Hyp Ref Expression
1 dmeqd.1 φA=B
2 dmeq A=BdomA=domB
3 1 2 syl φdomA=domB