Theorem eqimssi 3280

Description: Infer subclass relationship from equality. (Contributed by NM, 6-Jan-2007.)

Hypothesis

Ref	Expression
eqimssi.1	⊢ 𝐴 = 𝐵

Assertion

Ref	Expression
eqimssi	⊢ 𝐴 ⊆ 𝐵

Proof of Theorem eqimssi

Step	Hyp	Ref	Expression
1		ssid 3244	. 2 ⊢ 𝐴 ⊆ 𝐴
2		eqimssi.1	. 2 ⊢ 𝐴 = 𝐵
3	1, 2	sseqtri 3258	1 ⊢ 𝐴 ⊆ 𝐵

Syntax hints:   = wceq 1395   ⊆ wss 3197

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-11 1552  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-in 3203  df-ss 3210

This theorem is referenced by:  funi  5350  fpr  5821  elfzo1  10391  sumsplitdc  11943  isumlessdc  12007  nconstwlpolem0  16431