Theorem cnvxp 5049

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 5032	. . 3 |- `' { <. y , x >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( y e. A /\ x e. B ) }
2		ancom 266	. . . 4 |- ( ( y e. A /\ x e. B ) <-> ( x e. B /\ y e. A ) )
3	2	opabbii 4072	. . 3 |- { <. x , y >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( x e. B /\ y e. A ) }
4	1, 3	eqtri 2198	. 2 |- `' { <. y , x >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( x e. B /\ y e. A ) }
5		df-xp 4634	. . 3 |- ( A X. B ) = { <. y , x >. | ( y e. A /\ x e. B ) }
6	5	cnveqi 4804	. 2 |- `' ( A X. B ) = `' { <. y , x >. | ( y e. A /\ x e. B ) }
7		df-xp 4634	. 2 |- ( B X. A ) = { <. x , y >. | ( x e. B /\ y e. A ) }
8	4, 6, 7	3eqtr4i 2208	1 |- `' ( A X. B ) = ( B X. A )

Syntax hints:   /\ wa 104   = wceq 1353   e. wcel 2148   {copab 4065   X. cxp 4626   `'ccnv 4627

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-14 2151  ax-ext 2159  ax-sep 4123  ax-pow 4176  ax-pr 4211

This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-eu 2029  df-mo 2030  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-v 2741  df-un 3135  df-in 3137  df-ss 3144  df-pw 3579  df-sn 3600  df-pr 3601  df-op 3603  df-br 4006  df-opab 4067  df-xp 4634  df-rel 4635  df-cnv 4636

This theorem is referenced by:  xp0  5050  rnxpm  5060  rnxpss  5062  dminxp  5075  imainrect  5076  tposfo  6274  tposf  6275  xpider  6608  xpcomf1o  6827  pw1nct  14837