Theorem xp2dju 9602

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 5621	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8117	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4570	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2844	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5581	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9330	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2854	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1537   ∪ cun 3934  ∅c0 4291  {csn 4567  {cpr 4569   × cxp 5553  1_oc1o 8095  2_oc2o 8096   ⊔ cdju 9327

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 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-v 3496  df-dif 3939  df-un 3941  df-nul 4292  df-pr 4570  df-opab 5129  df-xp 5561  df-suc 6197  df-1o 8102  df-2o 8103  df-dju 9330

This theorem is referenced by:  pwdju1  9616  unctb  9627  infdjuabs  9628  ackbij1lem5  9646  fin56  9815