Theorem xp2dju 10171

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 5746	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8474	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4632	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2761	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5703	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9896	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2771	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1542   ∪ cun 3947  ∅c0 4323  {csn 4629  {cpr 4631   × cxp 5675  1_oc1o 8459  2_oc2o 8460   ⊔ cdju 9893

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 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3477  df-dif 3952  df-un 3954  df-nul 4324  df-pr 4632  df-opab 5212  df-xp 5683  df-suc 6371  df-1o 8466  df-2o 8467  df-dju 9896

This theorem is referenced by:  pwdju1  10185  unctb  10200  infdjuabs  10201  ackbij1lem5  10219  fin56  10388