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Theorem eqimss 2991
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 = BAB)

Proof of Theorem eqimss
StepHypRef Expression
1 eqss 2954 . 2 (A = B ↔ (AB BA))
21simplbi 259 1 (A = BAB)
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1242  wss 2911
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-in 2918  df-ss 2925
This theorem is referenced by:  eqimss2  2992  sspssr  3037  sspsstrir  3040  uneqin  3182  sssnr  3514  sssnm  3515  ssprr  3517  sstpr  3518  snsspw  3525  elpwuni  3731  disjeq2  3739  disjeq1  3742  pwne  3903  pwssunim  4011  poeq2  4027  seeq1  4060  seeq2  4061  trsucss  4125  onsucelsucr  4198  xp11m  4701  funeq  4862  fnresdm  4949  fssxp  4999  ffdm  5002  fcoi1  5011  fof  5047  dff1o2  5072  fvmptss2  5188  fvmptssdm  5196  fprg  5287  dff1o6  5357  tposeq  5800  nntri1  6006  frec2uzf1od  8819
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