Theorem cnvxp 4965

Description: The converse of a cross product. Exercise 11 of [Suppes] p. 67. (Contributed by NM, 14-Aug-1999.) (Proof shortened by Andrew Salmon, 27-Aug-2011.)

Assertion

Ref	Expression
cnvxp	|- `' ( A X. B ) = ( B X. A )

Proof of Theorem cnvxp

Dummy variables x y are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		cnvopab 4948	. . 3 |- `' { <. y , x >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( y e. A /\ x e. B ) }
2		ancom 264	. . . 4 |- ( ( y e. A /\ x e. B ) <-> ( x e. B /\ y e. A ) )
3	2	opabbii 4003	. . 3 |- { <. x , y >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( x e. B /\ y e. A ) }
4	1, 3	eqtri 2161	. 2 |- `' { <. y , x >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( x e. B /\ y e. A ) }
5		df-xp 4553	. . 3 |- ( A X. B ) = { <. y , x >. | ( y e. A /\ x e. B ) }
6	5	cnveqi 4722	. 2 |- `' ( A X. B ) = `' { <. y , x >. | ( y e. A /\ x e. B ) }
7		df-xp 4553	. 2 |- ( B X. A ) = { <. x , y >. | ( x e. B /\ y e. A ) }
8	4, 6, 7	3eqtr4i 2171	1 |- `' ( A X. B ) = ( B X. A )

Syntax hints:   /\ wa 103   = wceq 1332   e. wcel 1481   {copab 3996   X. cxp 4545   `'ccnv 4546

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-14 1493  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122  ax-sep 4054  ax-pow 4106  ax-pr 4139

This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1335  df-nf 1438  df-sb 1737  df-eu 2003  df-mo 2004  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-rex 2423  df-v 2691  df-un 3080  df-in 3082  df-ss 3089  df-pw 3517  df-sn 3538  df-pr 3539  df-op 3541  df-br 3938  df-opab 3998  df-xp 4553  df-rel 4554  df-cnv 4555

This theorem is referenced by:  xp0  4966  rnxpm  4976  rnxpss  4978  dminxp  4991  imainrect  4992  tposfo  6176  tposf  6177  xpider  6508  xpcomf1o  6727  pw1nct  13371