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Theorem dfss5 3115
 Description: Another definition of subclasshood. Similar to df-ss 2904, dfss 2905, and dfss1 3114. (Contributed by David Moews, 1-May-2017.)
Assertion
Ref Expression
dfss5 (ABA = (BA))

Proof of Theorem dfss5
StepHypRef Expression
1 dfss1 3114 . 2 (AB ↔ (BA) = A)
2 eqcom 2020 . 2 ((BA) = AA = (BA))
31, 2bitri 173 1 (ABA = (BA))
 Colors of variables: wff set class Syntax hints:   ↔ wb 98   = wceq 1226   ∩ cin 2889   ⊆ wss 2890 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-io 617  ax-5 1312  ax-7 1313  ax-gen 1314  ax-ie1 1359  ax-ie2 1360  ax-8 1372  ax-10 1373  ax-11 1374  ax-i12 1375  ax-bnd 1376  ax-4 1377  ax-17 1396  ax-i9 1400  ax-ial 1405  ax-i5r 1406  ax-ext 2000 This theorem depends on definitions:  df-bi 110  df-tru 1229  df-nf 1326  df-sb 1624  df-clab 2005  df-cleq 2011  df-clel 2014  df-nfc 2145  df-v 2533  df-in 2897  df-ss 2904 This theorem is referenced by: (None)
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