Theorem xp2dju 10259

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 5720	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8488	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4587	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2783	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5676	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9986	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2793	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

This proof depends on syntax axioms:   = wceq 1569   ∪ cun 4149  ∅c0 4278  {csn 4584  {cpr 4586   × cxp 5648  1_oc1o 8473  2_oc2o 8474   ⊔ cdju 9983

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-dif 4047  df-un 4151  df-nul 4279  df-pr 4587  df-opab 5167  df-xp 5656  df-suc 6368  df-1o 8480  df-2o 8481  df-dju 9986

This theorem is used by:  pwdju1  10273  unctb  10286  infdjuabs  10287  ackbij1lem5  10305  fin56  10475