CIS603 S03: Homework 3

1. Resolution Theorem Proving

Consider the following information:

1. Animals can outrun any animals that they can eat.
2. Carnivores eat other animals.
3. Outrunning is transitive: if x can outrun y and y can outrun z, then x can outrun z.
4. Lions eat zebras.
5. Zebras can outrun dogs.
6. Dogs are carnivores.

Solution: The following are the corresponding FOL expressions:

1. forall x, y:[ eats(x, y) --> outruns(x, y)]
2. forall x:[carnivorous(x) --> thereis y:[eats(x,y)]
3. forall x,y,z; [(outruns(x, y) and outruns(y, z) --> outruns(x, z)]
4. eats(Lions, Zebras)
5. outruns(Zebras, dogs)
6. carnivorous(Dogs)

1. not eats(x1, y1) or outruns(x1, y1)
2. not carnivorous(x2) or eats(x2, food(x2)
3. not outruns(x3, y2) or not outruns(y2, z1) or outruns(x3, z1)
4. eats(Lions, Zebras)
5. outruns(Zebras, dogs)
6. carnivorous(Dogs)

This can be shown using resolution as follows:

• (a) from 1 and 4 we get outruns( Lions, Zebras)
• (b) from (a) and 3 we get not outruns(Zebras, z1) or outruns(Lions, z1)
• (c) from (b) and 5 we get outruns(Lions, Dogs)
• (d) from 2 and 6 we get eats(Dogs, food(Dogs)
• (e) from (d) and 1 we get outruns(Dogs, food(Dogs))
• (f) from (c) and 3 we get not outruns(Sogs, z1) or outruns(Lions, z1)
• from (e) and (f) we get outruns(Lions, food(Lions))

2. First Order Logic

This problem can be solved in different ways, depending on the predicates you choose. Here is a sample solution.

1. Not all students take both History and Biology

   not forall x:[student(x) --> (takes(x, History) and takes(x, Biology))

2. Only one student failed history.

   thereis x forall y: [(student(x) and fails(x, History) and student(y) and fails(y, History)) --> x = y]

3. Only one student failed both History and Biology.

   thereis x forall y: [(student(x) and fails(x, History) and and fails(x, Biology) student(y) and fails(y, History) and fails(y, Biology)) --> x = y]

4. The best score in History was better than the best score in Biology.

   thereis x forall y: [score(x, History) > score(y, Biology)]

5. Every person who dislikes all vegetarians is smart.

   forall x forall y: [person(x) and (vegetarian(y) --> dislikes(x,y)) --> smart(y)]

6. No person likes a smart vegetarian.

   forall x forall y: [(person(x) and smart(y) and vegetarian(y)) --> not likes(x,y)

7. There is a woman who likes all men who are not vegetarian.

   thereis x foall y: [(woman(x) and man(y) and not vegetarian(y)) --> likes(x,y)]

8. There is a barber who shaves all men in town who are not vegetarian.

   thereis x forall y: [(barber(x) and man(y) and not shaves(y)) --> shaves(x,y)]

9. No person likes a professor unless the professor is smart.

   forall x forall y: [(person(x) and professor(y) and not smart(y)) --> not likes(x,y)]

10. Politicians can fool some of the people all of the time and they can fool all the people some of the time, but they can't fool all the people all of the time.

    forall x: [ politician(x) --> (thereis y forall t: [person(y) and fools(x,y,t)) and (thereis t forall y: [person(y) --> fools(x,y,t)]) and not ( forall t forall y: [person(y) --> fools(x,y,t)])]

3. Problem 9.4 from the textbook, p. 316.

The most general unifier (mgu) for the pairs of sentences are:

• (a) mgu(P(A,B,B),P(x,y,z))={x/A,y/B,z/B}.
• (b) mgu(Q(y,G(A,B)),Q(G(x,x),y))=FAIL since x cannot take on the value A and the value B at the same time.
• (c) mgu(Older(Father(y),y),Older(Father(x),John))={x/John,y/John}.
• (d) mgu(Knows(Father(y),y),Knows(x,y))=Fail since x cannot take on the value y and the value Father(y) at the same time, i.e., y cannot be Father(y).

Unification

What is the most general unifier for the expressions, where letters early in the alphabet are object constants, and letters late in the alphabet are variables.

p= f(x1, g(x2, x3), x2, b) and q = f(g(h(a, x5), x2), x1, h(a, x4), x4)

p = j(f(x, g(x,y)), h(z,y)) and q = j(z, h(f(u,v), f(a,b)))

Unifying the second arguments gives x1 = g(x2, x3), so we have x2 = h(a, x5) = x3.

The third argument gives x2 = h(a, x4). Combining this with the previous result for x2 gives x4 = x5. The fourth argument gives x4 = b. By composing these substitutions we obtain the final set of bindings:

x1 = g(h(a,b),h(a,b))
x2 = h(a,b)
x3 = h(a,b)
x4 = b
x5 = b

p = j(f(x, g(x,y)), h(z,y)) and q = j(z, h(f(u,v), f(a,b)))

Unifying the second arguments of h (which are the second arguments of j, we get y = f(a,b). Now we compose all these substitutions to get

u = x
v = g(x, f(a, b))
z = f(x, g(x,f(a,b)))
y = f(a,b)