Theorem cnvxp 4952

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 4935	. . 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 3990	. . 3 |- { <. x , y >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( x e. B /\ y e. A ) }
4	1, 3	eqtri 2158	. 2 |- `' { <. y , x >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( x e. B /\ y e. A ) }
5		df-xp 4540	. . 3 |- ( A X. B ) = { <. y , x >. | ( y e. A /\ x e. B ) }
6	5	cnveqi 4709	. 2 |- `' ( A X. B ) = `' { <. y , x >. | ( y e. A /\ x e. B ) }
7		df-xp 4540	. 2 |- ( B X. A ) = { <. x , y >. | ( x e. B /\ y e. A ) }
8	4, 6, 7	3eqtr4i 2168	1 |- `' ( A X. B ) = ( B X. A )

Syntax hints:   /\ wa 103   = wceq 1331   e. wcel 1480   {copab 3983   X. cxp 4532   `'ccnv 4533

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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-14 1492  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119  ax-sep 4041  ax-pow 4093  ax-pr 4126

This theorem depends on definitions:  df-bi 116  df-3an 964  df-tru 1334  df-nf 1437  df-sb 1736  df-eu 2000  df-mo 2001  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-ral 2419  df-rex 2420  df-v 2683  df-un 3070  df-in 3072  df-ss 3079  df-pw 3507  df-sn 3528  df-pr 3529  df-op 3531  df-br 3925  df-opab 3985  df-xp 4540  df-rel 4541  df-cnv 4542

This theorem is referenced by:  xp0  4953  rnxpm  4963  rnxpss  4965  dminxp  4978  imainrect  4979  tposfo  6161  tposf  6162  xpider  6493  xpcomf1o  6712