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Theorem xp2dju 7071
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 4596 . 2 (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2 df2o3 6327 . . . 4 2o = {∅, 1o}
3 df-pr 3534 . . . 4 {∅, 1o} = ({∅} ∪ {1o})
42, 3eqtri 2160 . . 3 2o = ({∅} ∪ {1o})
54xpeq1i 4559 . 2 (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6 df-dju 6923 . 2 (𝐴𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
71, 5, 63eqtr4i 2170 1 (2o × 𝐴) = (𝐴𝐴)
Colors of variables: wff set class
Syntax hints:   = wceq 1331  cun 3069  c0 3363  {csn 3527  {cpr 3528   × cxp 4537  1oc1o 6306  2oc2o 6307  cdju 6922
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 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121
This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-dif 3073  df-un 3075  df-nul 3364  df-pr 3534  df-opab 3990  df-suc 4293  df-xp 4545  df-1o 6313  df-2o 6314  df-dju 6923
This theorem is referenced by: (None)
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