Theorem xp2dju 10063

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 5681	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8388	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4574	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2754	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5637	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9789	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2764	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1541   ∪ cun 3895  ∅c0 4278  {csn 4571  {cpr 4573   × cxp 5609  1_oc1o 8373  2_oc2o 8374   ⊔ cdju 9786

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-v 3438  df-dif 3900  df-un 3902  df-nul 4279  df-pr 4574  df-opab 5149  df-xp 5617  df-suc 6307  df-1o 8380  df-2o 8381  df-dju 9789

This theorem is referenced by:  pwdju1  10077  unctb  10090  infdjuabs  10091  ackbij1lem5  10109  fin56  10279