Theorem xp2dju 10097

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 5695	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8410	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4565	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2763	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5651	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9823	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2773	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1547   ∪ cun 3888  ∅c0 4268  {csn 4562  {cpr 4564   × cxp 5623  1_oc1o 8395  2_oc2o 8396   ⊔ cdju 9820

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-v 3434  df-dif 3893  df-un 3895  df-nul 4269  df-pr 4565  df-opab 5142  df-xp 5631  df-suc 6323  df-1o 8402  df-2o 8403  df-dju 9823

This theorem is referenced by:  pwdju1  10111  unctb  10124  infdjuabs  10125  ackbij1lem5  10143  fin56  10313