Theorem xp2dju 7571

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 X. A ) = ( A ⊔ A )

Proof of Theorem xp2dju

Step	Hyp	Ref	Expression
1		xpundir 4832	. 2 |- ( ( { (/) } u. { 1o } ) X. A ) = ( ( { (/) } X. A ) u. ( { 1o } X. A ) )
2		df2o3 6702	. . . 4 |- 2o = { (/) , 1o }
3		df-pr 3716	. . . 4 |- { (/) , 1o } = ( { (/) } u. { 1o } )
4	2, 3	eqtri 2259	. . 3 |- 2o = ( { (/) } u. { 1o } )
5	4	xpeq1i 4794	. 2 |- ( 2o X. A ) = ( ( { (/) } u. { 1o } ) X. A )
6		df-dju 7378	. 2 |- ( A ⊔ A ) = ( ( { (/) } X. A ) u. ( { 1o } X. A ) )
7	1, 5, 6	3eqtr4i 2269	1 |- ( 2o X. A ) = ( A ⊔ A )

This proof depends on syntax axioms:   = wceq 1402   u. cun 3218   (/)c0 3520   {csn 3709   {cpr 3710   X. cxp 4772   1oc1o 6680   2oc2o 6681   ⊔ cdju 7377

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222  df-un 3224  df-nul 3521  df-pr 3716  df-opab 4193  df-suc 4516  df-xp 4780  df-1o 6687  df-2o 6688  df-dju 7378

This theorem is used by: (None)