Theorem xp2dju 10161

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 5733	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8462	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4593	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2786	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5689	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9888	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2796	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1570   ∪ cun 3904  ∅c0 4287  {csn 4590  {cpr 4592   × cxp 5661  1_oc1o 8447  2_oc2o 8448   ⊔ cdju 9885

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3457  df-dif 3909  df-un 3911  df-nul 4288  df-pr 4593  df-opab 5175  df-xp 5669  df-suc 6368  df-1o 8454  df-2o 8455  df-dju 9888

This theorem is referenced by:  pwdju1  10175  unctb  10188  infdjuabs  10189  ackbij1lem5  10207  fin56  10378