Theorem xp2dju 10191

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 5724	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8488	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4604	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2758	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5680	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9915	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2768	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1540   ∪ cun 3924  ∅c0 4308  {csn 4601  {cpr 4603   × cxp 5652  1_oc1o 8473  2_oc2o 8474   ⊔ cdju 9912

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-v 3461  df-dif 3929  df-un 3931  df-nul 4309  df-pr 4604  df-opab 5182  df-xp 5660  df-suc 6358  df-1o 8480  df-2o 8481  df-dju 9915

This theorem is referenced by:  pwdju1  10205  unctb  10218  infdjuabs  10219  ackbij1lem5  10237  fin56  10407