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Predicate logic proof

  1. Oct 5, 2008 #1
    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.
     
  2. jcsd
  3. Oct 9, 2008 #2
    If X may know Y, then by the first statement, there exists a Z such that X is a friend of Z and Z is a friend of Y. This means, by application of the second statement, that Y is a friend of Z and Z is a friend of X. So by the first statement, Y may know X
     
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