Theorem xp2dju 9932

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 5656	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8305	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4564	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2766	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5615	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9659	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2776	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1539   ∪ cun 3885  ∅c0 4256  {csn 4561  {cpr 4563   × cxp 5587  1_oc1o 8290  2_oc2o 8291   ⊔ cdju 9656

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-dif 3890  df-un 3892  df-nul 4257  df-pr 4564  df-opab 5137  df-xp 5595  df-suc 6272  df-1o 8297  df-2o 8298  df-dju 9659

This theorem is referenced by:  pwdju1  9946  unctb  9961  infdjuabs  9962  ackbij1lem5  9980  fin56  10149