Question

Given the following sentences:

1. All hounds howl at night.
2. Anyone who has any cats will not have any mice.
3. Light sleepers do not have anything which howls at night.
4. John has either a cat or a hound.
5. (Conclusion) If John is a light sleeper, then John does not have any mice.

Part 1:

1. Translate into propositional logic sentences
2. Convert the propositional sentences into Conjunctive Normal Form (CNF)

Answer 1

a.)

1. ∀x(Hound(x) → Howl(x))
2. ∀x∀y(Have(x, y) ∧ Cat(y) → ¬∃z(Have(x, z) ∧ Mouse(z)))
3. ∀x(LS(x) → ¬∃y(Have(x, y) ∧ Howl(y)))
4. ∃x(Have(John, x) ∧ (Cat(x) ∨ Hound(x)))
5. LS(John) → ¬∃z(Have(John, z) ∧ Mouse(z))

b.)

1. ¬Hound(x) ∨ Howl(x)
2. ¬Have(x, y) ∨ ¬Cat(y) ∨ ¬Have(x, z) ∨ ¬Mouse(z)
3. ¬LS(x) ∨ ¬Have(x, y) ∨ ¬Howl(y)
4. (a) Have(John, a)
(b) Cat(a) ∨ Hound(a)
5. (a) LS(John)
(b) Have(John, b)
(c) Mouse(b)


we proceed to prove the conclusion by resolution using the above clauses. Each result clause
is numbered; the numbers of its parent clauses are shown at the right-hand side.
6. Cat(a) ∨ Howl(a)                        [1, 4(b)]
7. ¬Have(x, y) ∨ ¬Cat(y) ∨ ¬Have(x, b)     [2, 5(c)]
8. ¬Have(John, y) ∨ ¬Cat(y)                [7, 5(b)]
9. ¬Have(John, a) ∨ Howl(a)                [6, 8]
10. Howl(a)                                [4(a), 9]
11. ¬LS(x) ∨ ¬Have(x, a)                   [3, 10]
12. ¬LS(John)                              [4(a), 11]
13.                                        [5(a), 12]