Theorem eqimss2 3192

Description: Equality implies the subclass relation. (Contributed by NM, 23-Nov-2003.)

Assertion

Ref	Expression
eqimss2	⊢ (𝐵 = 𝐴 → 𝐴 ⊆ 𝐵)

Proof of Theorem eqimss2

Step	Hyp	Ref	Expression
1		eqimss 3191	. 2 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
2	1	eqcoms 2167	1 ⊢ (𝐵 = 𝐴 → 𝐴 ⊆ 𝐵)

Syntax hints:   → wi 4   = wceq 1342   ⊆ wss 3111

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 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-11 1493  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522  ax-ext 2146

This theorem depends on definitions:  df-bi 116  df-nf 1448  df-sb 1750  df-clab 2151  df-cleq 2157  df-clel 2160  df-in 3117  df-ss 3124

This theorem is referenced by:  disjeq2  3957  disjeq1  3960  poeq2  4272  seeq1  4311  seeq2  4312  dmcoeq  4870  xp11m  5036  funeq  5202  fconst3m  5698  tposeq  6206  undifdcss  6879  ennnfonelemk  12270  ennnfonelemss  12280  qnnen  12301  topgele  12568  topontopn  12576  txdis  12818