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Theorem xp2dju 7152
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 4645 . 2 (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2 df2o3 6379 . . . 4 2o = {∅, 1o}
3 df-pr 3568 . . . 4 {∅, 1o} = ({∅} ∪ {1o})
42, 3eqtri 2178 . . 3 2o = ({∅} ∪ {1o})
54xpeq1i 4608 . 2 (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6 df-dju 6984 . 2 (𝐴𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
71, 5, 63eqtr4i 2188 1 (2o × 𝐴) = (𝐴𝐴)
Colors of variables: wff set class
Syntax hints:   = wceq 1335  cun 3100  c0 3395  {csn 3561  {cpr 3562   × cxp 4586  1oc1o 6358  2oc2o 6359  cdju 6983
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 604  ax-in2 605  ax-io 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139
This theorem depends on definitions:  df-bi 116  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-v 2714  df-dif 3104  df-un 3106  df-nul 3396  df-pr 3568  df-opab 4028  df-suc 4333  df-xp 4594  df-1o 6365  df-2o 6366  df-dju 6984
This theorem is referenced by: (None)
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