Theorem eqimss2 3197

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 3196	. 2 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
2	1	eqcoms 2168	1 ⊢ (𝐵 = 𝐴 → 𝐴 ⊆ 𝐵)

Syntax hints:   → wi 4   = 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:  disjeq2  3963  disjeq1  3966  poeq2  4278  seeq1  4317  seeq2  4318  dmcoeq  4876  xp11m  5042  funeq  5208  fconst3m  5704  tposeq  6215  undifdcss  6888  ennnfonelemk  12333  ennnfonelemss  12343  qnnen  12364  topgele  12667  topontopn  12675  txdis  12917