Theorem xp2dju 10246

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 5769	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8530	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4651	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2768	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5726	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9970	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2778	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1537   ∪ cun 3974  ∅c0 4352  {csn 4648  {cpr 4650   × cxp 5698  1_oc1o 8515  2_oc2o 8516   ⊔ cdju 9967

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-v 3490  df-dif 3979  df-un 3981  df-nul 4353  df-pr 4651  df-opab 5229  df-xp 5706  df-suc 6401  df-1o 8522  df-2o 8523  df-dju 9970

This theorem is referenced by:  pwdju1  10260  unctb  10273  infdjuabs  10274  ackbij1lem5  10292  fin56  10462