Theorem xp2dju 10215

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) = (𝐴 ⊔ 𝐴)

Proof of Theorem xp2dju

Step	Hyp	Ref	Expression
1		xpundir 5758	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8513	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4634	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2763	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5715	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9939	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2773	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1537   ∪ cun 3961  ∅c0 4339  {csn 4631  {cpr 4633   × cxp 5687  1_oc1o 8498  2_oc2o 8499   ⊔ cdju 9936

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-v 3480  df-dif 3966  df-un 3968  df-nul 4340  df-pr 4634  df-opab 5211  df-xp 5695  df-suc 6392  df-1o 8505  df-2o 8506  df-dju 9939

This theorem is referenced by:  pwdju1  10229  unctb  10242  infdjuabs  10243  ackbij1lem5  10261  fin56  10431