dean,
does this help? I don't know pilog well but was just playing around

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: (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 <deangwillia...@gmail.com> wrote:
> (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 <deangwillia...@gmail.com> wrote:
>>
>> Oops....let me try this
>>
>> (-> @X) in place of @X in the lisp clause
>>
>>
>> On 27 November 2016 at 16:38, dean <deangwillia...@gmail.com> 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 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 do this.
>>>
>>>
>>> : (prove (goal '( (^ @X (- 4 2)) (equal @A 4) (equal @B 2)
>>> )))
>>> -> ((@X . 2) (@A . 4) (@B . 2))
>>> : (prove (goal '( (^ @X (- @A @B)) (equal @A 4) (equal @B 2)
>>> )))
>>> -> NIL
>>>
>>>
>>> On 27 November 2016 at 08:46, dean <deangwillia...@gmail.com> wrote:
>>>>
>>>> Ok I'll keep trying and thank you for the pointers.
>>>> Best Regardsd
>>>> Dean
>>>>
>>>> On 27 November 2016 at 07:33, Alexander Burger <a...@software-lab.de>
>>>> wrote:
>>>>>
>>>>> Hi Dean,
>>>>>
>>>>> > #(prove (goal '(equal 3 @X) ))
>>>>>
>>>>> 'goal' needs a list of clauses:
>>>>>
>>>>> : (prove (goal '((equal 3 @X))))
>>>>> -> ((@X . 3))
>>>>>
>>>>>
>>>>> > #: (prove (goal '( (equal 3 @X) (member @X (1 2 4)) )))
>>>>> > #-> NIL
>>>>> > #: (prove (goal '( (equal 3 @X) (member @X (1 2 3)) )))
>>>>> > #-> ((@X . 3))
>>>>>
>>>>> OK
>>>>>
>>>>>
>>>>> > #(prove (goal '(
>>>>> > # (equal @Profit (- @Sales @Cogs))
>>>>>
>>>>> Did you define a '-' predicate?
>>>>>
>>>>> ♪♫ Alex
>>>>> --
>>>>> UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
>>>>
>>>>
>>>
>>
>
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