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Metamath Proof Explorer
Description: Infer subclass relationship from equality. (Contributed by NM, 7-Jan-2007)
|
|
Ref |
Expression |
|
Hypothesis |
eqimssi.1 |
⊢ 𝐴 = 𝐵 |
|
Assertion |
eqimss2i |
⊢ 𝐵 ⊆ 𝐴 |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
eqimssi.1 |
⊢ 𝐴 = 𝐵 |
| 2 |
|
ssid |
⊢ 𝐵 ⊆ 𝐵 |
| 3 |
2 1
|
sseqtrri |
⊢ 𝐵 ⊆ 𝐴 |