Theorem eqimss2 3280

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 3279	. 2 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
2	1	eqcoms 2232	1 ⊢ (𝐵 = 𝐴 → 𝐴 ⊆ 𝐵)

Syntax hints:   → wi 4   = wceq 1395   ⊆ wss 3198

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 3204  df-ss 3211

This theorem is referenced by:  ifpprsnssdc  3777  disjeq2  4066  disjeq1  4069  poeq2  4395  seeq1  4434  seeq2  4435  dmcoeq  5003  xp11m  5173  funeq  5344  fconst3m  5868  tposeq  6408  undifdcss  7108  nninfctlemfo  12601  ennnfonelemk  13011  ennnfonelemss  13021  qnnen  13042  imasaddfnlemg  13387  topgele  14743  topontopn  14751  txdis  14991  edgstruct  15905