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Theorem ssequn2 3089
Description: A relationship between subclass and union. (Contributed by NM, 13-Jun-1994.)
Assertion
Ref Expression
ssequn2 (AB ↔ (BA) = B)

Proof of Theorem ssequn2
StepHypRef Expression
1 ssequn1 3086 . 2 (AB ↔ (AB) = B)
2 uncom 3060 . . 3 (AB) = (BA)
32eqeq1i 2025 . 2 ((AB) = B ↔ (BA) = B)
41, 3bitri 173 1 (AB ↔ (BA) = B)
Colors of variables: wff set class
Syntax hints:  wb 98   = wceq 1226  cun 2888  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-un 2895  df-in 2897  df-ss 2904
This theorem is referenced by:  unabs  3140  pwssunim  3991  pwundifss  3992  oneluni  4114  relresfld  4770  relcoi1  4772  fsnunf  5283
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