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Re: solving for pilog variables

Joe Yes that's cracked it! I can see you've given values to @A and @B before solving for @C but in Prolog I wasn't aware that the order mattered...There again the calculation is being done in Lisp. As a result of Alex's response this morning I added more parenthesis and that seemed to solve what I

Re: solving for pilog variables

Hi Dean On Sun, Nov 27, 2016 at 05:42:21PM +, dean wrote: > (prove (goal '( (^ @X (- (-> @A) (-> @B) )) (equal @A 4) (equal @B 2) ))) > -> NIL > > -> wasn't the "one" in this case I'm not sure I understand the problem, but the most natural way for a diff predicate is perhaps : (be -

Re: solving for pilog variables

dean, does this help? I don't know pilog well but was just playing around : (prove (goal '( (equal @A 5) (equal @B 2) (^ @C (- (-> @A) (-> @B) ) -> ((@A . 5) (@B . 2) (@C . 3)) On Sun, Nov 27, 2016 at 12:42 PM, dean wrote: > (prove (goal '( (^ @X (- (-> @A)

Re: solving for pilog variables

(prove (goal '( (^ @X (- (-> @A) (-> @B) )) (equal @A 4) (equal @B 2) ))) -> NIL -> wasn't the "one" in this case On 27 November 2016 at 17:34, dean wrote: > Oopslet me try this > > (-> @X) in place of @X in the lisp clause > > > On 27 November 2016 at 16:38,

Re: solving for pilog variables

Oopslet me try this (-> @X) in place of @X in the lisp clause On 27 November 2016 at 16:38, dean wrote: > In preparation to do a predicate 'minus' I thought I see how to use lisp > '-' within pilog. > The first statement works re getting a 100% lisp calculation

Re: solving for pilog variables

In preparation to do a predicate 'minus' I thought I see how to use lisp '-' within pilog. The first statement works re getting a 100% lisp calculation out to pilog but I think I need to pass in pilog variables and apply - to them...unless pilog has a -. Many apologies if I should know how to

Re: solving for pilog variables

Ok I'll keep trying and thank you for the pointers. Best Regardsd Dean On 27 November 2016 at 07:33, Alexander Burger wrote: > Hi Dean, > > > #(prove (goal '(equal 3 @X) )) > > 'goal' needs a list of clauses: > >: (prove (goal '((equal 3 @X >-> ((@X . 3)) > >