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Theorem xp2dju 7216
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 4685 . 2 (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2 df2o3 6433 . . . 4 2o = {∅, 1o}
3 df-pr 3601 . . . 4 {∅, 1o} = ({∅} ∪ {1o})
42, 3eqtri 2198 . . 3 2o = ({∅} ∪ {1o})
54xpeq1i 4648 . 2 (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6 df-dju 7039 . 2 (𝐴𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
71, 5, 63eqtr4i 2208 1 (2o × 𝐴) = (𝐴𝐴)
Colors of variables: wff set class
Syntax hints:   = wceq 1353  cun 3129  c0 3424  {csn 3594  {cpr 3595   × cxp 4626  1oc1o 6412  2oc2o 6413  cdju 7038
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 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-dif 3133  df-un 3135  df-nul 3425  df-pr 3601  df-opab 4067  df-suc 4373  df-xp 4634  df-1o 6419  df-2o 6420  df-dju 7039
This theorem is referenced by: (None)
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