Theorem xp2dju 9790

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 5618	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8217	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4544	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2765	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5577	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9517	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2775	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1543   ∪ cun 3864  ∅c0 4237  {csn 4541  {cpr 4543   × cxp 5549  1_oc1o 8195  2_oc2o 8196   ⊔ cdju 9514

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-8 2112  ax-9 2120  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-tru 1546  df-fal 1556  df-ex 1788  df-sb 2071  df-clab 2715  df-cleq 2729  df-clel 2816  df-v 3410  df-dif 3869  df-un 3871  df-nul 4238  df-pr 4544  df-opab 5116  df-xp 5557  df-suc 6219  df-1o 8202  df-2o 8203  df-dju 9517

This theorem is referenced by:  pwdju1  9804  unctb  9819  infdjuabs  9820  ackbij1lem5  9838  fin56  10007