Theorem xp2dju 10127

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 5713	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8439	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4582	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2784	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5669	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9853	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2794	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1559   ∪ cun 3900  ∅c0 4283  {csn 4579  {cpr 4581   × cxp 5641  1_oc1o 8424  2_oc2o 8425   ⊔ cdju 9850

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-fal 1572  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3455  df-dif 3905  df-un 3907  df-nul 4284  df-pr 4582  df-opab 5160  df-xp 5649  df-suc 6347  df-1o 8431  df-2o 8432  df-dju 9853

This theorem is referenced by:  pwdju1  10141  unctb  10154  infdjuabs  10155  ackbij1lem5  10173  fin56  10344