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Theorem xp2dju 7035
 Description: Two times a cardinal number. Exercise 4.56(g) of [Mendelson] p. 258. (Contributed by NM, 27-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
Assertion
Ref Expression
xp2dju (2o × 𝐴) = (𝐴𝐴)

Proof of Theorem xp2dju
StepHypRef Expression
1 xpundir 4564 . 2 (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2 df2o3 6293 . . . 4 2o = {∅, 1o}
3 df-pr 3502 . . . 4 {∅, 1o} = ({∅} ∪ {1o})
42, 3eqtri 2136 . . 3 2o = ({∅} ∪ {1o})
54xpeq1i 4527 . 2 (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6 df-dju 6889 . 2 (𝐴𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
71, 5, 63eqtr4i 2146 1 (2o × 𝐴) = (𝐴𝐴)
 Colors of variables: wff set class Syntax hints:   = wceq 1314   ∪ cun 3037  ∅c0 3331  {csn 3495  {cpr 3496   × cxp 4505  1oc1o 6272  2oc2o 6273   ⊔ cdju 6888 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 586  ax-in2 587  ax-io 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-bndl 1469  ax-4 1470  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498  ax-ext 2097 This theorem depends on definitions:  df-bi 116  df-tru 1317  df-nf 1420  df-sb 1719  df-clab 2102  df-cleq 2108  df-clel 2111  df-nfc 2245  df-v 2660  df-dif 3041  df-un 3043  df-nul 3332  df-pr 3502  df-opab 3958  df-suc 4261  df-xp 4513  df-1o 6279  df-2o 6280  df-dju 6889 This theorem is referenced by: (None)
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