Theorem xp2dju 10156

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 5729	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8457	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4594	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2792	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5685	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9883	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2802	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1567   ∪ cun 3911  ∅c0 4294  {csn 4591  {cpr 4593   × cxp 5657  1_oc1o 8442  2_oc2o 8443   ⊔ cdju 9880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-v 3465  df-dif 3916  df-un 3918  df-nul 4295  df-pr 4594  df-opab 5175  df-xp 5665  df-suc 6363  df-1o 8449  df-2o 8450  df-dju 9883

This theorem is referenced by:  pwdju1  10170  unctb  10183  infdjuabs  10184  ackbij1lem5  10202  fin56  10373