Theorem xp2dju 10167

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 5743	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8470	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4630	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2760	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5701	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9892	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2770	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1541   ∪ cun 3945  ∅c0 4321  {csn 4627  {cpr 4629   × cxp 5673  1_oc1o 8455  2_oc2o 8456   ⊔ cdju 9889

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-fal 1554  df-ex 1782  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-v 3476  df-dif 3950  df-un 3952  df-nul 4322  df-pr 4630  df-opab 5210  df-xp 5681  df-suc 6367  df-1o 8462  df-2o 8463  df-dju 9892

This theorem is referenced by:  pwdju1  10181  unctb  10196  infdjuabs  10197  ackbij1lem5  10215  fin56  10384