Theorem xp2dju 10183

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 5729	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8467	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4590	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2785	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5685	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9910	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2795	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

This proof depends on syntax axioms:   = wceq 1570   ∪ cun 3900  ∅c0 4282  {csn 4587  {cpr 4589   × cxp 5657  1_oc1o 8452  2_oc2o 8453   ⊔ cdju 9907

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3455  df-dif 3905  df-un 3907  df-nul 4283  df-pr 4590  df-opab 5172  df-xp 5665  df-suc 6367  df-1o 8459  df-2o 8460  df-dju 9910

This theorem is used by:  pwdju1  10197  unctb  10210  infdjuabs  10211  ackbij1lem5  10229  fin56  10399