Metamath Proof Explorer


Theorem eqimss2i

Description: Infer subclass relationship from equality. (Contributed by NM, 7-Jan-2007)

Ref Expression
Hypothesis eqimssi.1 𝐴 = 𝐵
Assertion eqimss2i 𝐵𝐴

Proof

Step Hyp Ref Expression
1 eqimssi.1 𝐴 = 𝐵
2 ssid 𝐵𝐵
3 2 1 sseqtrri 𝐵𝐴