Theorem xp2dju 9595

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 5614	. 2 ⊢ (({∅} ∪ {1o}) × 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
2		df2o3 8110	. . . 4 ⊢ 2o = {∅, 1o}
3		df-pr 4563	. . . 4 ⊢ {∅, 1o} = ({∅} ∪ {1o})
4	2, 3	eqtri 2843	. . 3 ⊢ 2o = ({∅} ∪ {1o})
5	4	xpeq1i 5574	. 2 ⊢ (2o × 𝐴) = (({∅} ∪ {1o}) × 𝐴)
6		df-dju 9323	. 2 ⊢ (𝐴 ⊔ 𝐴) = (({∅} × 𝐴) ∪ ({1o} × 𝐴))
7	1, 5, 6	3eqtr4i 2853	1 ⊢ (2o × 𝐴) = (𝐴 ⊔ 𝐴)

Syntax hints:   = wceq 1536   ∪ cun 3927  ∅c0 4284  {csn 4560  {cpr 4562   × cxp 5546  1_oc1o 8088  2_oc2o 8089   ⊔ cdju 9320

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-ext 2792

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1539  df-ex 1780  df-nf 1784  df-sb 2069  df-clab 2799  df-cleq 2813  df-clel 2892  df-nfc 2962  df-v 3493  df-dif 3932  df-un 3934  df-nul 4285  df-pr 4563  df-opab 5122  df-xp 5554  df-suc 6190  df-1o 8095  df-2o 8096  df-dju 9323

This theorem is referenced by:  pwdju1  9609  unctb  9620  infdjuabs  9621  ackbij1lem5  9639  fin56  9808