Theorem xp2dju 10294

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 5728	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8522	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4591	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2786	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5684	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 10019	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2796	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

This proof depends on syntax axioms:   = wceq 1570   ∪ cun 4152  ∅c0 4281  {csn 4588  {cpr 4590   × cxp 5656  1_oc1o 8507  2_oc2o 8508   ⊔ cdju 10016

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-dif 4050  df-un 4154  df-nul 4282  df-pr 4591  df-opab 5174  df-xp 5664  df-suc 6394  df-1o 8514  df-2o 8515  df-dju 10019

This theorem is used by:  pwdju1  10308  unctb  10321  infdjuabs  10322  ackbij1lem5  10340  fin56  10510