Theorem xp2dju 10099

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 5701	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8413	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4570	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2759	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5657	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9825	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2769	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1542   ∪ cun 3887  ∅c0 4273  {csn 4567  {cpr 4569   × cxp 5629  1_oc1o 8398  2_oc2o 8399   ⊔ cdju 9822

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-v 3431  df-dif 3892  df-un 3894  df-nul 4274  df-pr 4570  df-opab 5148  df-xp 5637  df-suc 6329  df-1o 8405  df-2o 8406  df-dju 9825

This theorem is referenced by:  pwdju1  10113  unctb  10126  infdjuabs  10127  ackbij1lem5  10145  fin56  10315