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 ( 𝜑 → dom 𝐴 = dom 𝐵 )

Proof

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