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Theorem xp2dju 7571
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 4832 . 2 (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2 df2o3 6702 . . . 4 2o = {∅, 1o}
3 df-pr 3716 . . . 4 {∅, 1o} = ({∅} ∪ {1o})
42, 3eqtri 2259 . . 3 2o = ({∅} ∪ {1o})
54xpeq1i 4794 . 2 (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6 df-dju 7378 . 2 (𝐴𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
71, 5, 63eqtr4i 2269 1 (2o × 𝐴) = (𝐴𝐴)
Colors of variables:    wff set class
This proof depends on syntax axioms:   = wceq 1402  cun 3218  c0 3520  {csn 3709  {cpr 3710   × cxp 4772  1oc1o 6680  2oc2o 6681  cdju 7377
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220
This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222  df-un 3224  df-nul 3521  df-pr 3716  df-opab 4193  df-suc 4516  df-xp 4780  df-1o 6687  df-2o 6688  df-dju 7378
This theorem is used by: (None)
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