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## RE: [saxon] Value.inverse

 RE: [saxon] Value.inverse From: Michael Kay - 2004-03-30 15:08:00 ```Sorry, I don't understand the problem. AA, which is what this code is doing. Michael Kay > -----Original Message----- > From: saxon-help-admin@... > [mailto:saxon-help-admin@...] On Behalf Of > Colin Paul Adams > Sent: 30 March 2004 14:19 > To: saxon-help@... > Subject: [saxon] Value.inverse > > In both 7.8 and 7.9.1 you have a routine inverse: > > /** > * Return the inverse of a relational operator, so that > "a op b" can be > * rewritten as "b inverse(op) a" > */ > > But there appears to be a bug here. > > a < b yields the same result as b >= a > > yet you have > as the inverse of <, and are likewise consistent with > all the other non-symmetrical operators. > -- > Colin Paul Adams > Preston Lancashire > > > ------------------------------------------------------- > This SF.Net email is sponsored by: IBM Linux Tutorials > Free Linux tutorial presented by Daniel Robbins, President and CEO of > GenToo technologies. Learn everything from fundamentals to system > administration.http://ads.osdn.com/?ad_id=1470&alloc_id=3638&op=click > _______________________________________________ > saxon-help mailing list > saxon-help@... > https://lists.sourceforge.net/lists/listinfo/saxon-help > ```

 [saxon] Value.inverse From: Colin Paul Adams - 2004-03-30 14:18:40 ```In both 7.8 and 7.9.1 you have a routine inverse: /** * Return the inverse of a relational operator, so that "a op b" can be * rewritten as "b inverse(op) a" */ But there appears to be a bug here. a < b yields the same result as b >= a yet you have > as the inverse of <, and are likewise consistent with all the other non-symmetrical operators. -- Colin Paul Adams Preston Lancashire ```
 RE: [saxon] Value.inverse From: Michael Kay - 2004-03-30 15:08:00 ```Sorry, I don't understand the problem. AA, which is what this code is doing. Michael Kay > -----Original Message----- > From: saxon-help-admin@... > [mailto:saxon-help-admin@...] On Behalf Of > Colin Paul Adams > Sent: 30 March 2004 14:19 > To: saxon-help@... > Subject: [saxon] Value.inverse > > In both 7.8 and 7.9.1 you have a routine inverse: > > /** > * Return the inverse of a relational operator, so that > "a op b" can be > * rewritten as "b inverse(op) a" > */ > > But there appears to be a bug here. > > a < b yields the same result as b >= a > > yet you have > as the inverse of <, and are likewise consistent with > all the other non-symmetrical operators. > -- > Colin Paul Adams > Preston Lancashire > > > ------------------------------------------------------- > This SF.Net email is sponsored by: IBM Linux Tutorials > Free Linux tutorial presented by Daniel Robbins, President and CEO of > GenToo technologies. Learn everything from fundamentals to system > administration.http://ads.osdn.com/?ad_id=1470&alloc_id=3638&op=click > _______________________________________________ > saxon-help mailing list > saxon-help@... > https://lists.sourceforge.net/lists/listinfo/saxon-help > ```
 Re: [saxon] Value.inverse From: Colin Paul Adams - 2004-03-30 15:22:52 ```>>>>> "Michael" == Michael Kay writes: Michael> Sorry, I don't understand the problem. A legitimately be rewritten as B> A, which is what this code is doing. Sorry, I made a mistake. -- Colin Paul Adams Preston Lancashire ```