Theorem cnvxp 5022

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 5005	. . 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 4049	. . 3 |- { <. x , y >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( x e. B /\ y e. A ) }
4	1, 3	eqtri 2186	. 2 |- `' { <. y , x >. | ( y e. A /\ x e. B ) } = { <. x , y >. | ( x e. B /\ y e. A ) }
5		df-xp 4610	. . 3 |- ( A X. B ) = { <. y , x >. | ( y e. A /\ x e. B ) }
6	5	cnveqi 4779	. 2 |- `' ( A X. B ) = `' { <. y , x >. | ( y e. A /\ x e. B ) }
7		df-xp 4610	. 2 |- ( B X. A ) = { <. x , y >. | ( x e. B /\ y e. A ) }
8	4, 6, 7	3eqtr4i 2196	1 |- `' ( A X. B ) = ( B X. A )

Syntax hints:   /\ wa 103   = wceq 1343   e. wcel 2136   {copab 4042   X. cxp 4602   `'ccnv 4603

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 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-14 2139  ax-ext 2147  ax-sep 4100  ax-pow 4153  ax-pr 4187

This theorem depends on definitions:  df-bi 116  df-3an 970  df-tru 1346  df-nf 1449  df-sb 1751  df-eu 2017  df-mo 2018  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-rex 2450  df-v 2728  df-un 3120  df-in 3122  df-ss 3129  df-pw 3561  df-sn 3582  df-pr 3583  df-op 3585  df-br 3983  df-opab 4044  df-xp 4610  df-rel 4611  df-cnv 4612

This theorem is referenced by:  xp0  5023  rnxpm  5033  rnxpss  5035  dminxp  5048  imainrect  5049  tposfo  6239  tposf  6240  xpider  6572  xpcomf1o  6791  pw1nct  13883