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CSE/CIS 607 Mathematical Basis of Computing
Practice Exam II
Question 1: Is the following argument valid? Explain.
Given: Clark comes from a place called Krypton or Lex retired. Conclusion: If Clark does not come from a place called Krypton, then the moon is made of cheese or Lex retired.
Answer: Let K be “Clark comes from a place called Krypton”. Let L be “Lex retired”. Let C be “The moon is made of cheese”. The argument has the form:
premise: K ∨ L conclusion: ¬K ⊃ (C ∨ L)
By Gentzen’s method, we can see that if we try to falsify the conclusion whole holding the premise true, it must be that ¬K is true, but C ∨ L is false. Therefore, K is false and L is false. Therefore the K ∨ L is false, which contradicts that K ∨ L is true. Thus the conclusion must be true if the premise is true. Hence, the argument is valid.
Question 2: Give a formal tableau proof that shows that the following tableau is valid. g1. (P ⊃ (Q ⊃ R)) ⊃ ((P ∧ ¬R) ⊃ ¬Q) given
Answer:
g1. (P ⊃ (Q ⊃ R)) ⊃ ((P ∧ ¬R) ⊃ ¬Q) given a1. P ⊃ (Q ⊃ R) g1 if-split g2. (P ∧ ¬R) ⊃ ¬Q g1 if-split a2. P ∧ ¬R g2 if-split g3. ¬Q g2 if-split a3. P a3 ∧-split a4. ¬R a3 ∧-split a5. False ∨ (True ⊃ (Q ⊃ R)) a1, a3 AA a6. Q ⊃ R a5 rew a7. (Q ⊃ False) ∨ ¬True a6, a4 AA a8. ¬Q a7 rew g4. ¬False ∧ True a8, g3 AG g5. True g4 rew
Question 3: What’s wrong with the following proposed inference rule? Let P be a common subformula of assertion A and goal G, occurring at least once in each of these formulas. Then
From goal G 1 and goal G 2 derive goal G 1 [P 7 → F alse] ∨ G 2 [P 7 → T rue]
(This is similar to, but not the same as, GG-resolution.)
Answer: The following solution is due fundamentally to Robert C. Grant, and has the virtue of conforming to the polarity strategy.
g1. ¬P given g2. P ∧ False g2 if-split
This tableau is not valid since the tableau’s corresponding formula is not valid:
¬P ∨ (P ∧ False)
is not valid. The formula is not valid since it can be falsified by assigning T to P. However, if we apply the proposed inference rule to the tableau, we obtain
g1. ¬P given g2. P ∧ False g2 if-split g3. ¬False ∨ (True ∧ False) g1, g g4. True g3 rew
tableau proof if is valid.)
a1. (J ∧ K) ⊃ L given a2. (J ⊃ L) ⊃ M given a3. ¬K ∨ N given g1. K ⊃ (M ∧ N ) given
Answer: The tableau is valid. The tableau would not be valid if all of its goals can be falsified while the assertions are true. Suppose K ⊃ (M ∧ N ) is false and the assertions are true. Then K is true and M ∧ N is false. Since ¬K ∨ N is true, but ¬K is false, N is true. Since, M ∧ N is false and N is true, M is false. Since (J ⊃ L) ⊃ M is true, but M is false, J ⊃ L is false. Since J ⊃ L is false, J is true and L is false. Since, (J ∧ K) ⊃ L is true, but L is false, J ∧ K is false. Since J ∧ K is false, and J is true, K is false. But K is true and we have a contradiction. Therefore if we suppose the tableau is not valid, a contradiction results.
Question 7: Give a formal tableau proof that shows that the following tableau is valid. a1. (A ∨ B) ⊃ (C ∧ D) given a2. ¬C given g1. ¬B given
Answer: a1. (A ∨ B) ⊃ (C ∧ D) given a2. ¬C given g1. ¬B given a3. ((A ∨ B) ⊃ (False ∧ D)) ∨ ¬True a1, a2 AA a4. ¬A ∧ ¬B a3 rew a5. ¬B a4 ∧-split g2. ¬False ∧ True a5, g2 AG g3. True g2 rew
Question 8: Give a formal tableau proof that shows that the following tableau is
valid. a1. Q ⊃ ¬P given a2. P ⊃ Q given g1. ¬P given
Answer: a1. Q ⊃ ¬P given a2. P ⊃ Q given g1. ¬P given g2. ¬(True ⊃ Q) ∧ ¬False g1, a2 GA g3. ¬Q g2 rew g4. ¬(True ⊃ ¬P ) ∧ ¬False g3, a1 GA g5. P g4 rew g6. ¬False ∧ True g1, g6 GG g7. True g6 rew
Question 9: What’s wrong with the following proposed inference rule? Let P be a common subformula of assertion A 1 and assertion A 2. Then
From assertion A 1 and assertion A 2 derive assertion A 1 [P 7 → F alse] ∧ A 2 [P 7 → T rue]
Question 10: Consider:
Assertion 1: If Alice doesn’t catch the bus, then Bob gambles away his money or Bob leaves town.
Assertion 2: If Alice catches the bus and Bob does not leave town, then Bob gambles away his money.
Assertion 3: If Bob gambles away his money, then Bob leaves town.
Goal: Bob leaves town.
Answer: a1. (P ∧ Q) ⊃ R given a2. P given a3. ¬Q ⊃ (¬P ∨ R) given g1. R given a4. ¬P ∨ ¬Q ∨ R a1 rew a5. ¬P ∨ Q ∨ R a3 rew a6. False ∨ (¬True ∨ ¬Q ∨ R a2, a4 AA a7. ¬Q ∨ R a6 rew a8. False ∨ (¬True ∨ Q ∨ R a2, a5 AA a9. Q ∨ R a2, a5 AA a10. False ∨ (¬True ∨ R) a9, a7 AA a11. R a10 rew g2. ¬False ∧ True a11, g2 AG g3. True g2 rew
Question 13: What’s wrong with the following proposed inference rule? Let P be a common subformula of assertion A and goal G, occurring at least once in each of these formulas. Then
From assertion A and goal G derive goal A[P 7 → F alse] ∧ G[P 7 → T rue]
(This is similar to, but not the same as, AG-resolution.)
Question 14: This question is optional - there will not be a question like it on the upcoming exam. BUT, making an effort to do the question will help you reinforce your understanding of propositional logic. The diagram below denotes a Toffoli gate. P, Q and R are Boolean-valued. The expression PQ + R in the diagram denotes the Boolean-value of the propositional formula (P ∧ Q) ⊕ R. (⊕ denotes exclusive-or.)
P Q R
P Q PQ + R
Notice that if the value of R is fixed at true, then the Toffoli gate acts as a NAND gate:
P Q true
P Q PQ + true
Show how to make the Toffoli gate act as an XOR gate. XOR(A,B) is equivalent to the propositional formula A ⊕ B
Question 15: Consider:
Assertion 1: If Alice sends an encrypted text to Bob, then Bob will be safe and Bob doesn’t leave town.
Assertion 2: If Alice does not send an encrypted text to Bob and Bob leaves town, then Bob will not be safe.
Assertion 3: Bob doesn’t leave town or Bob is safe.
Goal: Bob does not leave town.
Let A stand for “Alice sends an encrypted text to Bob.” Let B stand for “Bob will be safe.” Let L stand for “Bob leaves town.”
(1) Express the assertions and goal in our notation for propositional logic.
(2) Does the goal follow from the assertion? (Explain.)
Question 16: Is the following tableau valid? If it is, give a formal tableau proof. If
Question 18: This question is optional - there will not be a question like it on the upcoming exam. BUT, making an effort to do the question will help you reinforce your understanding of propositional logic. The diagram below denotes a Fredkin gate. P, Q and R are Boolean-valued. The expression PQ + PR in the diagram denotes the Boolean-value of the propositional formula (¬ P ∧ Q) ⊕ (P ∧ R). (⊕ denotes exclusive-or.)
P Q R
P PQ + PR PR + PQ
Notice that if the value of R is fixed at true, then the Fredkin gate acts as an OR gate:
P Q true
P PQ + P P + PQ
Show how to make the Fredkin gate act as an AND gate.
Question 19: Use the unification algorithm to determine whether there is a most
general unifying substitution that solves the equation
k(f (x, u, a), f (g(y), y, v)) = k(f (g(y), y, v), f (w, h(z), z))
In the preceding equation, a is a constant, and x, y, z, u, v and w are variables.
Answers: k(f (x, u, a), f (g(y), y, v)) = k(f (g(y), y, v), f (w, h(z), z))
f (x, u, a) = f (g(y), y, v) f (g(y), y, v) = f (w, h(z), z)
x = g(y)
u = y a = v g(y) = w y = h(z) v = z
x = g(h(z)) u = h(z) v = a w = g(h(z)) y = h(z) z = a
x = g(h(a)) u = h(a) v = a w = g(h(a)) y = h(a) z = a
k(f (g(h(a)), h(a), a), f (g(h(a)), h(a), a)) = k(f (g(h(a)), h(a), a), f (g(h(a)), h(a), a))
Question 20: Use the unification algorithm to determine whether there is a most
general unifying substitution that solves the equation
j(g(x, u, w), g(f (y), y, v)) = j(g(f (y), y, v), g(x, k(z), z))
In the preceding equation, x, y, z, u, v and w are variables.
Answer: j(g(x, u, w), g(f (y), y, v)) = j(g(f (y), y, v), g(x, k(z), z))
g(x, u, w) = g(f (y), y, v) g(f (y), y, v) = g(x, k(z), z)
Question 22: Let Ψ be a formula in which the subformula P ∨ ¬P has exactly one occurrence, and that occurrence has negative polarity in Ψ. Is the following formula valid? If it is valid, explain why. If it is not valid give a (simple) counterexample.
(Ψ[P ∨ ¬P ] ∧ Q) ⊃ (Ψ[P ∧ ¬P ] ∧ Q)
Question 23: Find a most general unifier of the two terms
g(g(w, g(b, z)), w) and g(g(y, g(b, u)), g(u, z)).
Answer: g(g(w, g(b, z)), w) = g(g(y, g(b, u)), g(u, z))
g(w, g(b, z)) = g(y, g(b, u)) w = g(u, z)
w = y g(b, z) = g(b, u) w = g(u, z)
w = y g(b, z) = g(b, u) y = g(u, z)
w = y b = b z = u y = g(u, z)
w = y z = u y = g(u, z)
w = y z = u
y = g(u, u)
w = g(u, u) z = u y = g(u, u)
g(g(g(u,u),g(b,u)),g(u,u)) = g(g(g(u,u),g(b,u)),g(u,u))
Question 24: Find a most general unifier of the two terms
g(x, y, f (x), z, u) and g(a, x, z, f (x), h(v, z)).
Answer: g(x, y, f (x), z, u) = g(a, x, z, f (x), h(v, z))
x = a y = x f (x) = z z = f (x) u = h(v, z)
x = a y = a f (a) = z z = f (a) u = h(v, z)
x = a y = a z = f (a) z = f (a) u = h(v, z)
x = a y = a
Question 26: Find a most general unifier of the two terms
f (h(y, x, g(v, x))), f (h(v, x, g(g(u, u), f (u)))).
Answer: f (h(y, x, g(v, x))) = f (h(v, x, g(g(u, u), f (u))))
h(y, x, g(v, x)) = h(v, x, g(g(u, u), f (u)))
y = v x = x g(v, x) = g(g(u, u), f (u))
y = v g(v, x) = g(g(u, u), f (u))
y = v v = g(u, u) x = f (u)
y = g(u, u) v = g(u, u) x = f (u)
f (h(g(u, u), f (u), g(g(u, u), f (u)))) = f (h(g(u, u), f (u), g(g(u, u), f (u))))
Question 27: Intentionally omitted
Question 28: Is the following tableau valid? If it is, give a formal tableau proof. If it is not, explain why. g1. (P ⊃ (Q ∧ R)) ⊃ ((P ⊃ Q) ⊃ (P ⊃ R)).
Answer: The tableau is valid.
g1. (P ⊃ (Q ∧ R)) ⊃ ((P ⊃ Q) ⊃ (P ⊃ R)) given g2. True g1 rew
(Seriously!)
Question 29: Is the following tableau valid? If it is, give a formal tableau proof. If it is not, explain why.
g1. ((P ⊃ Q) ⊃ (P ⊃ R)) ⊃ (P ⊃ (Q ∧ R)).
Answer: The tableau is not valid because the goal can be falsified while all of the assertions are true: Assign T to P and R and F to Q.
Question 30: Give a formal tableau proof that shows that the following tableau is valid. a1. P ⊃ (Q ≡ P ) given a2. ¬P ⊃ R given g1. R ∨ (Q ≡ P ) given
(Hint: You do not need to use the equivalence rule, and you do not need to eliminate the equivalances with rewrites.)
Answer: a1. P ⊃ (Q ≡ P ) given a2. ¬P ⊃ R given g1. R ∨ (Q ≡ P ) given a3. (¬False ⊃ R) ∨ (True ⊃ (Q ≡ P )) a2, a1 AA a4. R ∨ (Q ≡ P ) a3 rew g2. ¬False ∧ True a4, g1 AG g3. True g2 rew
a1. P ⊃ (Q ⊃ R) a2. (R ∧ S) ⊃ ¬U a3. ¬V ⊃ (S ∧ U ) g1. Q ⊃ (¬P ∨ ¬R)
Answer: The tableau is not valid. Let T be assigned to P , Q, R, S, and V , and let F be assigned to U. Then the goal is false, and the assertions are true.
Question 35: Suppose that subformula P occurs at least once in each of the formulas A and B. Explain why the following tableau is valid:
g1. (A ∧ B) ⊃ (A[P 7 → True] ∨ B[P 7 → False])
Answer:
g1. (A ∧ B) ⊃ (A[P 7 → T rue] ∨ B[P 7 → F alse]) given a1. A ∧ B g1 if-split g2. A[P 7 → True] ∨ B[P 7 → False] g1 if-split a2. A a1 ∧-split a3. B a1 ∧-split a4. A[P 7 → True] ∨ B[P 7 → False] a2, a3 AA g3. ¬False ∧ True a4, g2 AG g4. True g3 rew
