Solutions

Solutions

p(X) <- q(X).
p(Y) <- r(Y).
q(a).
r(b).

1. Give a model of KB where D={fred,mary}. You must specify phi and pi.

   One model is where a, b and c all denote fred (i.e., phi(a)=fred, phi(b)=fred, and phi(c)=fred) and all relations are true (i.e., pi(p)(fred)=true, pi(p)(mary)=true, pi(q)(fred)=true, pi(q)(mary)=true, pi(r)(fred)=true, pi(r)(mary)=true).

   Another model is where a denotes fred and b and c denote mary (i.e., phi(a)=fred, phi(b)=mary, and phi(c)=mary) and the relations are as follows: pi(p)(fred)=true, pi(p)(mary)=true, pi(q)(fred)=true, pi(q)(mary)=false, pi(r)(fred)=false, pi(r)(mary)=true).

   Another model is where a, b and c all denote fred (i.e., phi(a)=fred, phi(b)=fred, and phi(c)=fred) and the relations are as follows: pi(p)(fred)=true, pi(p)(mary)=false, pi(q)(fred)=true, pi(q)(mary)=false, pi(r)(fred)=true, pi(r)(mary)=false).

2. Give an interpretation with the same domain that isn't a model of KB. You must specify phi and pi.

   We can select an arbitrary clause and falsify it, filling the other atoms arbitrarily. For example, any interpretation with a denoting fred (i.e., with phi(a)=fred) and pi(q)(fred)=true falsifies the third clause and isn't a model.

   Similarly any interpretation with pi(p)(fred)=false and pi(p)(fred)=true falsifies the first clause, and so isn't a model, independently of the values of ]phi or the other values of pi.

   So one interpretation that isn't a model is where a denotes fred and b and c denote mary (i.e., phi(a)=fred, phi(b)=mary, and phi(c)=mary) and the relations are as follows: pi(p)(fred)=false, pi(p)(mary)=false, pi(q)(fred)=false, pi(q)(mary)=true, pi(r)(fred)=true, pi(r)(mary)=false). Every clause in the knowledge base is false in this interpretation.

3. Give three atoms that are logical consequences of the knowledge base.

   Any two of q(a), r(b), p(a), p(b)

4. Give three atoms that are not logical consequences of the knowledge base.

   Any two of r(a), q(b), r(c), q(c), p(c).

Solutions