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Theorem xp2dju 10167
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 5743 . 2 (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2 df2o3 8470 . . . 4 2o = {∅, 1o}
3 df-pr 4630 . . . 4 {∅, 1o} = ({∅} ∪ {1o})
42, 3eqtri 2760 . . 3 2o = ({∅} ∪ {1o})
54xpeq1i 5701 . 2 (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6 df-dju 9892 . 2 (𝐴𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
71, 5, 63eqtr4i 2770 1 (2o × 𝐴) = (𝐴𝐴)
Colors of variables: wff setvar class
Syntax hints:   = wceq 1541  cun 3945  c0 4321  {csn 4627  {cpr 4629   × cxp 5673  1oc1o 8455  2oc2o 8456  cdju 9889
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-fal 1554  df-ex 1782  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-v 3476  df-dif 3950  df-un 3952  df-nul 4322  df-pr 4630  df-opab 5210  df-xp 5681  df-suc 6367  df-1o 8462  df-2o 8463  df-dju 9892
This theorem is referenced by:  pwdju1  10181  unctb  10196  infdjuabs  10197  ackbij1lem5  10215  fin56  10384
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