To correct the resolution, we have to change the name of variable y in either first clause or second clause.. Question 2: What is wrong with the following argument: •Men are widely distr
Trang 1Tutorial 6 PREDICATE LOGIC and STRUCTURED KNOWLEDGE
Question 1:
Wrong in:
¬gt(5, 2) ∨ ¬succ(y, 2) combines with 6¬gt(x, y) ∨ ¬gt(y, z) ∨ gt(x, z) by the substitution 5/x, y/z
This combination is wrong because the variable y in each clause is different to each other
To correct the resolution, we have to change the name of variable y in either first clause or second clause
Question 2:
What is wrong with the following argument:
•Men are widely distributed over the earth
•Socrates is a man
•Therefore, Socrates is widely distributed over the earth
How should the facts represented by these sentences be represented in logic so that this problem does not arise?
Solution
The conclusion is false because the property "widely distributed over the earth" is appli-cable over a set (group of men) Not to each individual in the set
To avoid that wrong conclusion, we can represent these facts like this:
man(Socrates)
widelyDistributed(men,earth)
Question 3:
Consider the following axioms:
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
Use resolution with predicate logic to affirm the following conclusion:
If John is a light sleeper, then John does not have any mice
Solution
Trang 22 ∀x∀y(HAV E(x, y) ∧ CAT (y) → ¬∃z(HAV E(x, z) ∧ M OU SE(z)))
3 ∀x(LS(x) → ¬∃y(HAV E(x, y) ∧ HOW L(y)))
4 ∃x(HAV E(John, x) ∧ ((CAT (x) ∨ HOU N D(x)) ∧ (¬CAT (x) ∨ ¬HOU N D(x))) Conclusion: LS(John) → ¬∃z(HAV E(John, z) ∧ M OU SE(z))
Convert to clause form
1.¬HOU N D(x) ∨ HOW L(x)
2.¬HAV E(x, y) ∨ ¬CAT (y) ∨ ¬HAV E(x, z) ∨ ¬M OU SE(z)
3.¬LS(x) ∨ ¬HAV E(x, y) ∨ ¬HOW L(y)
4a.HAV E(John, a)
4b.CAT (a) ∨ HOU N D(a)
4c ¬CAT (a) ∨ ¬HOU N D(a)
Negation of conclusion
5a.LS(John)
5b.HAV E(John, b)
5c.M OU SE(b)
Resolution
5c b/z 2 6.¬HAV E(x, y) ∨ ¬CAT (y) ∨ ¬HAV E(x, b)
5b John/x 6 7.¬HAV E(John, y) ∨ ¬CAT (y)
4b a/y 7 8.¬HAV E(John, a) ∨ HOU N D(a)
3 a/y 9 10.¬LS(x) ∨ ¬HAV E(x, a)
4a John/x 10 11.¬LS(John)
Trang 3(P ( (()) ) ) Double negation
(P ( ((P)) ) ) Iteration
(P ( ( Q (P) ) ) ) Insertion
(P ( ( Q (P) ) ) ) Insertion
b: ( ( ( (S (F)) ) ( (H (F)) ) ) ((S H (F))) )
( ( ((S H (F))) ) ((S H (F))) ) Iteration
( ( S H (F) ) ((S H (F))) ) Double negation
( ( S H (F) (F) ) ((S H (F))) ) Iteration
( ( S (F) ((H (F))) ) ((S H (F))) ) Double negation
( ( ((S (F))) ((H (F))) ) ((S H (F))) ) Double negation
c: ( (Q ((P))) ((Q (P)) ((Q))) )
( () ((Q (P)) ((Q))) ((Q (P)) ((Q))) ) Iteration
( () ((Q (P)) Q) ((Q (P)) ((Q))) ) Double negation
( () (((P)) Q) ((Q (P)) ((Q))) ) Deiteration
( () ( P Q) ((Q (P)) ((Q))) ) Double negation
( ( (P Q) ) (P Q) ((Q (P)) ((Q))) ) iteration
( ( (Q ((P))) ) (P Q) ((Q (P)) ((Q))) ) Double negation
Trang 4d: ( (P (Q)) ((P) (R)) (((P Q) ((P) R))) )
( () ((P)) ((Q)) ((P) (R))) Insertion
( (((P)) ((Q))) ((P)) ((Q)) ((P) (R)) ) Iteration
( (P Q) ((P)) ((Q)) ((R)) ((P) (R)) ) Double negation
( (P Q) ((P)) ((Q) ((P)) ) ((P) (R)) ) Iteration
( (P Q) ((P)) ((Q) P) ((P) (R)) ) Double negation
( (P Q) ((Q) P) ((P) ((P) (R))) ((P) (R)) ) Iteration
( (P Q) ((Q) P) ( (P) R ) ((P) (R)) ) Deiteration + Double negation
( (P (Q)) ((P) (R)) (P Q) ((P) R) ) Just reorder
( (P (Q)) ((P) (R)) (( (P Q) ((P) R) )) ) Double negation
Question 5:
Convert the following sentence into predicate logic, existential graph (EG) and conceptual graph (CG):
If a cat is on a mat, then it is a happy pet
Solution
Prediate logic:
∀x∀y : cat(x) ∧ mat(y) ∧ on(x, y) → happyP et(x)
¬(∃x∃y : cat(x) ∧ mat(y) ∧ on(x, y) ∧ ¬happyP et(x)
Another possible solution:
∀x∀y∀z((cat(x) ∧ mat(y) ∧ on(x, y)) → (pet(x) ∧ happy(z) ∧ haveattr(x, z)))