Re: pilog in ersatz

2012-07-10 Thread Alexander Burger
Hi Christophe, Yes. This is an infinite recursion. ... So it seems it's not a syntax problem, but a conceptual one. I thought about @x and @y being the same person, but it shouldn't. Another hint? I'm not a Prolog expert, but fact is that if the first three clauses (be bigger (me her))

Re: pilog in ersatz

2012-07-10 Thread Christophe Gragnic
On Mon, Jul 9, 2012 at 11:12 PM, Doug Snead semaphore_2...@yahoo.com wrote: Hmmm. Where do alix and sw originate? I do not understand that. Sorry for the confusion, I was doing my tests with real names in my family but thought afterwards I should use names that could be understood by anybody.

Re: pilog in ersatz

2012-07-10 Thread Alexander Burger
On Tue, Jul 10, 2012 at 09:56:26AM +0200, Alexander Burger wrote: (be bigger (@x @y) (isbigger @x @z) (isbigger @z @y)) No, I think the last rule must be (be bigger (@x @y) (isbigger @x @z) (bigger @z @y)) Cheers, - Alex -- UNSUBSCRIBE:

Re: pilog in ersatz

2012-07-10 Thread Christophe Gragnic
On Tue, Jul 10, 2012 at 9:56 AM, Alexander Burger a...@software-lab.de wrote: don't find a match, the fourth one (be bigger (@x @y) (bigger @x @z) (bigger @z @y)) will always match and recurse infinitely. Is it because it always match or because to know if it would succeed, it must test

Re: pilog in ersatz

2012-07-09 Thread Doug Snead
The pilog trace can be helpful in these situations. : (be bigger (me her)) : (be bigger (her son)) : (be bigger (son daughter)) : (be bigger (@x @y) (bigger @x @z) (bigger @z @y)) List the assertions you want to trace before the clause to be proved, in this case bigger : : (? bigger (bigger @x

Re: pilog in ersatz

2012-07-09 Thread Doug Snead
--- On Sun, 7/8/12, Christophe Gragnic christophegrag...@gmail.com wrote: : (be bigger (me her)) - bigger : (be bigger (her son)) - bigger : (be bigger (son daughter)) - bigger : (be bigger (@x @y) (bigger @x @z) (bigger @z @y)) - bigger : bigger - NIL : (? (bigger @x meily)) @x=alix