Theorem eqimss 3323

Description: Equality implies the subclass relation. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)

Assertion

Ref	Expression
eqimss	|- (A = B -> A C_ B)

Proof of Theorem eqimss

Step	Hyp	Ref	Expression
1		eqss 3287	. 2 |- (A = B <-> (A C_ B /\ B C_ A))
2	1	simplbi 446	1 |- (A = B -> A C_ B)

Syntax hints:   -> wi 4   = wceq 1642   C_ wss 3257

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-ss 3259

This theorem is referenced by:  eqimss2  3324  sspss  3368  uneqin  3506  uneqdifeq  3638  pwpw0  3855  sssn  3864  eqsn  3867  snsspw  3877  pwsnALT  3882  unissint  3950  elpwuni  4053  iotassuni  4355  dmxpss  5052  xp11  5056  dmsnopss  5067  fnresdm  5192  fssxp  5232  ffdm  5234  fof  5269  dff1o2  5291  dff1o6  5475  fvmptss  5705  fvmptss2  5725  qsss  5986  enprmaplem6  6081