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Theorem xp2dju 7192
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 4668 . 2 (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2 df2o3 6409 . . . 4 2o = {∅, 1o}
3 df-pr 3590 . . . 4 {∅, 1o} = ({∅} ∪ {1o})
42, 3eqtri 2191 . . 3 2o = ({∅} ∪ {1o})
54xpeq1i 4631 . 2 (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6 df-dju 7015 . 2 (𝐴𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
71, 5, 63eqtr4i 2201 1 (2o × 𝐴) = (𝐴𝐴)
Colors of variables: wff set class
Syntax hints:   = wceq 1348  cun 3119  c0 3414  {csn 3583  {cpr 3584   × cxp 4609  1oc1o 6388  2oc2o 6389  cdju 7014
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 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-dif 3123  df-un 3125  df-nul 3415  df-pr 3590  df-opab 4051  df-suc 4356  df-xp 4617  df-1o 6395  df-2o 6396  df-dju 7015
This theorem is referenced by: (None)
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