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Theorem xp2dju 7473
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 4789 . 2 (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2 df2o3 6640 . . . 4 2o = {∅, 1o}
3 df-pr 3680 . . . 4 {∅, 1o} = ({∅} ∪ {1o})
42, 3eqtri 2252 . . 3 2o = ({∅} ∪ {1o})
54xpeq1i 4751 . 2 (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6 df-dju 7280 . 2 (𝐴𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
71, 5, 63eqtr4i 2262 1 (2o × 𝐴) = (𝐴𝐴)
Colors of variables: wff set class
Syntax hints:   = wceq 1398  cun 3199  c0 3496  {csn 3673  {cpr 3674   × cxp 4729  1oc1o 6618  2oc2o 6619  cdju 7279
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213
This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-v 2805  df-dif 3203  df-un 3205  df-nul 3497  df-pr 3680  df-opab 4156  df-suc 4474  df-xp 4737  df-1o 6625  df-2o 6626  df-dju 7280
This theorem is referenced by: (None)
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