Theorem eqimssi 3198

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 3162	. 2 ⊢ 𝐴 ⊆ 𝐴
2		eqimssi.1	. 2 ⊢ 𝐴 = 𝐵
3	1, 2	sseqtri 3176	1 ⊢ 𝐴 ⊆ 𝐵

Syntax hints:   = wceq 1343   ⊆ wss 3116

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-11 1494  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-in 3122  df-ss 3129

This theorem is referenced by:  funi  5220  fpr  5667  elfzo1  10125  sumsplitdc  11373  isumlessdc  11437  nconstwlpolem0  13941