I have a question on my assignment that I'm having a great deal of trouble with. In this question we will consider a simplified model of the “People you may know” application in Facebook. The basic idea is that if two people have a common friend, then they may know each other. We will also take into account the fact in Facebook the relation “friend” is symmetric, that is, if X is a friend of Y , then Y is a friend of X. Thus, as a problem description P, we have the following two predicate logic sentences: 8X8Y [(9Z(friend(X, Z) ^ friend(Z, Y ))) $ may know(X, Y )] 8X8Y [friend(X, Y ) ! friend(Y,X)] Our goal is to show, using resolution refutation, that the following sentence g is implied by P: 8X8Y [may know(X, Y ) ! may know(Y,X)]. 8 = universal quantifier 9 = existential quantifier $ = biconditional ! = implication I have been able to derive the following clauses (this is for a logical programming class): c1: may_know(X,Y) :- friend(X,T), friend(T,Y) c2: friend(X,f(X,Y,T)) :- may_know(X,Y) c3: friend(f(X,Y,T),Y) :- may_know(X,Y) c4: friend(Y,X) :- friend(X,Y) c5: may_know(d,e) :- c6: :- may_know(e,d) I've attempted linear refutation which won't work. The question goes on to advise that you prove g from p "using your normal, mathematical, reasoning. Then, try to re-create the steps of your proof using resolution on your clauses." This is where I'm stuck, I'm not sure how I can prove g from p using either my clauses or my ordinary mathematical reasoning. Any help with this would be greatly appreciated.