Theorem eqimss2 3279

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 3278	. 2 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
2	1	eqcoms 2232	1 ⊢ (𝐵 = 𝐴 → 𝐴 ⊆ 𝐵)

Syntax hints:   → wi 4   = 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:  ifpprsnssdc  3774  disjeq2  4063  disjeq1  4066  poeq2  4391  seeq1  4430  seeq2  4431  dmcoeq  4997  xp11m  5167  funeq  5338  fconst3m  5862  tposeq  6399  undifdcss  7096  nninfctlemfo  12576  ennnfonelemk  12986  ennnfonelemss  12996  qnnen  13017  imasaddfnlemg  13362  topgele  14718  topontopn  14726  txdis  14966  edgstruct  15879