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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  3515  sssnm  3516  ssprr  3518  sstpr  3519  snsspw  3526  elpwuni  3732  disjeq2  3740  disjeq1  3743  pwne  3904  pwssunim  4012  poeq2  4028  seeq1  4061  seeq2  4062  trsucss  4126  onsucelsucr  4199  xp11m  4702  funeq  4864  fnresdm  4951  fssxp  5001  ffdm  5004  fcoi1  5013  fof  5049  dff1o2  5074  fvmptss2  5190  fvmptssdm  5198  fprg  5289  dff1o6  5359  tposeq  5803  nntri1  6013  frec2uzf1od  8873
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