Theorem eqimss 3324

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 3288	. 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 3258

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 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260

This theorem is referenced by:  eqimss2  3325  sspss  3369  uneqin  3507  uneqdifeq  3639  pwpw0  3856  sssn  3865  eqsn  3868  snsspw  3878  pwsnALT  3883  unissint  3951  elpwuni  4054  iotassuni  4356  dmxpss  5053  xp11  5057  dmsnopss  5068  fnresdm  5193  fssxp  5233  ffdm  5235  fof  5270  dff1o2  5292  dff1o6  5476  fvmptss  5706  fvmptss2  5726  qsss  5987  enprmaplem6  6082