Distributivity check in Prolog -


suppose have equivalence relation eq, , multiple binary operators o_1, o_2, ... o_n. want find out operations distribute on others. assuming have knowledge base can determine whether 2 expressions equivalent, simple solution enter possible queries :

(for left-distributivity)

?- eq(o_1(z,o_1(x,y)),o_1(o_1(z,x),o_1(z,y))).  ?- eq(o_1(z,o_2(x,y)),o_2(o_1(z,x),o_1(z,y))).  ?- eq(o_1(z,o_3(x,y)),o_3(o_1(z,x),o_1(z,y))).   ...  ?- eq(o_2(z,o_2(x,y)),o_2(o_2(z,x),o_2(z,y))).  ?- eq(o_2(z,o_3(x,y)),o_3(o_2(z,x),o_2(z,y))).   ...  ?- eq(o_n(z,o_n(x,y)),o_n(o_n(z,x),o_n(z,y))).

but there must better ways this. starters, i'd define predicate left_dist such left_dist(o_m,o_k) generate corresponding query me. thought i'd using call, in

left_dist(o_m,o_k) :-    eq(call(o_m,z,call(o_k,x,y)),call(o_k,call(o_m,z,x),call(o_m,z,y))).

but nested calls not work reasons outlined in this question, , guess it's not way approach prolog programming either.

so question : how can define left_dist, or otherwise simplify queries above, in prolog?

left distributivity means x,y,z: x*(y+z)=(x*y)+(x*z)

now not mention concrete domains, assume relations know them already. assumes add_3 , mult_3 terminate.

not_left_dist(mult_3, add_3) :-    call(add_3, y, z, yz),    call(mult_3, x, yz, lhs),    call(mult_3, x, y, xy),    call(mult_3, x, z, xz),    call(add_3, xy, xz, rhs),    neq(lhs, rhs).  left_dist(mult_3, add_3) :-    iwhen(ground(mult_3+add_3), \+ not_left_dist(mult_3, add_3) ).

this uses iwhen/2.


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