Theorem xp2dju 9863

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 5647	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8282	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4561	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2766	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5606	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9590	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2776	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1539   ∪ cun 3881  ∅c0 4253  {csn 4558  {cpr 4560   × cxp 5578  1_oc1o 8260  2_oc2o 8261   ⊔ cdju 9587

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3424  df-dif 3886  df-un 3888  df-nul 4254  df-pr 4561  df-opab 5133  df-xp 5586  df-suc 6257  df-1o 8267  df-2o 8268  df-dju 9590

This theorem is referenced by:  pwdju1  9877  unctb  9892  infdjuabs  9893  ackbij1lem5  9911  fin56  10080