Finding the difference in consequences between two logic programs

Hello,

Given two logic programs LP1 and LP2, I would like to generate
consequences of LP1 that are not consequences of LP2. Is there an
efficient way of finding these consequences? A basic solution is
quite simple to express:

lp1_and_not_lp2(X) :- lp1(X), \+ lp2(X).

but as the programs become more complex, the inefficiency of this
approach becomes a problem.

It seems to be quite a general problem - is there any theory I should
know about, and is another setting (e.g. CLP / answer set programming)
more suited than Prolog?

In my case, LP1 and LP2 are closely related (to get LP2 just add/
remove a few clauses from LP1) - does this help make the problem
easier to solve?

Thanks for any pointers!
Zoubidoo


A simple example:

% LP1
lp1(X) :- lp1_p(X).
lp1_p(a).
lp1_p(b).

% LP2 Removed p(b) and added p(c)
lp2(X) :- lp2_p(X).
lp2_p(a).
lp2_p(c).

lp1_and_not_lp2(X) :- lp1(X), \+ lp2(X).

The only solution here is X = b.

On Jul 1, 7:20*am, Zoubidoo <Stanc...@gmail.com> wrote:
> Hello,
>
> Given two logic programs LP1 and LP2, I would like to generate
> consequences of LP1 that are not consequences of LP2. *Is there an
> efficient way of finding these consequences? *A basic solution is
> quite simple to express:
>
> lp1_and_not_lp2(X) :- lp1(X), \+ lp2(X).
>
> but as the programs become more complex, the inefficiency of this
> approach becomes a problem.
>
> It seems to be quite a general problem - is there any theory I should
> know about, and is another setting (e.g. CLP / answer set programming)
> more suited than Prolog?
>
> In my case, LP1 and LP2 are closely related (to get LP2 just add/
> remove a few clauses from LP1) - does this help make the problem
> easier to solve?
>
> Thanks for any pointers!
> Zoubidoo
>
> A simple example:
>
> % LP1
> lp1(X) :- lp1_p(X).
> lp1_p(a).
> lp1_p(b).
>
> % LP2 * Removed p(b) and added p(c)
> lp2(X) :- lp2_p(X).
> lp2_p(a).
> lp2_p(c).
>
> lp1_and_not_lp2(X) :- lp1(X), \+ lp2(X).
>
> The only solution here is X = b.

Without "special knowledge" of predicates
lp1/1 and lp2/1, the opportunities for
improving on your basic solution seem
limited.

One alternative, if the solutions to lp1/1
are finite, is to form a list of these L
via findall/3, then process that list to
knock out those which solve lp2/1 as well.

Such an approach might be especially
attractive if the comparison were to
involve more than two predicates.

regards, chip