Advances in Artificial Intelligence Theory - MICAI - Mexican ...

Prov(p(d);α(d)) - Prov(X1;Y1) - Prov(X1,α(a),c=a;p(a),Y1)&Prov(X1;c=a,Y1) -
- Prov(X1;c=a,Y1) - Prov(p(d),α(c);c=a,c=b,p(c),α(d)) - □.
The second node is given by matching with C1 and is the first meta-predicate in the
body of C1 since T(p/α,x/c’:X,p(x); α(x),Y) is true. The third is given by applying C2
with y=a. By applying A1 to the first meta-predicate Prov(X1,p(a),c=a;p(a),Y1) and
Prov(X1,p(b),c=b,c=a;p(a),Y1) are obtained. Both are true by P1 and by C3 with a=b
false, respectively, and the fourth node is given. Similarly the fifth node is obtained
by applying C2 with y=b. The last is given by A1 and P1.
It is noted that {x=a,x=b}←p(x) is also proved in Example 7. Then the following
two circumscription meta-rules corresponding to α(x)≡x=a and α(x)≡x=b are used.
Prov(X,p(x);x=a,Y) ← Prov(X,p(x),c=a;p(c),x=a,Y),
Prov(X,p(x);x=b,Y) ← Prov(X,p(x),c'=b;p(c'),x=b,Y).
Similarly the well-known solution, i.e., for all x and all y {(x=a←p(x)),
(y=b←p(y))}, is also proved in MC by using the above two circumscription metarules
with the following meta-rule for the query and constants, d and d’, not in MC.
Prov(X;Y) ← Prov(X,p(d),p(d’);d=a,d’=b,Y).
Example 8 (The priority and variable predicates) With A, S, E, p1, and p2 for
adult, student, employed, abnormal1, and abnormal2, respectively, Let T be
p1(x)←S(x)&E(x), {E(x),p2(x)}←A(x), A(x)←S(x), S(m)← and let the query be
~E(m), where m (Mary) is a constant. The corresponding meta-rules are,
respectively,
Prov(X,S(x),E(x);p1(x),Y) ←, (4)
Prov(X,A(x);E(x),p2(x),Y) ←, (5)
Prov(X,S(x);A(x),Y) ←, (6)
Prov(X;S(m),Y) ←, (7)
Prov(X;Y) ← Prov(X,E(m);Y). (8)
The circumscription meta-rules for p1 and p2 are, respectively,
Prov(X,p1(x);α1(x),Y)
←Prov(X,p1(x),α1(c);p1(c),α1(x),Y)&Prov(X,p1(x),S(c’),E(c’);α1(c’),α1(x),Y), (9)
Prov(X,p2(x);α2(x),Y)
←Prov(X,p2(x),α2(d);p2(d),α2(x),Y)&Prov(X,p2(x),A(d’);E(d’),α2(d’),α2(x),Y).(10)