Artificial Intelligence: A Modern Approach
3rd Edition
ISBN: 9780136042594
Author: Stuart Russell, Peter Norvig
Publisher: Prentice Hall
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Expert Solution & Answer
Chapter 7, Problem 18E
a.
Explanation of Solution
Truth table
- A simple truth table has eight rows...
b.
Explanation of Solution
Results
- For the left-hand side:
(Food ⇒ Party) ∨ (Drinks ⇒ Party)
(¬Food ∨ Party) ∨ (¬Drinks ∨ Party)
(¬Food ∨ Party ∨ ¬Drinks ∨ Party)
(¬Food ∨ ¬Drinks ∨ Party)
- For the right-ha...
c.
Explanation of Solution
Resolution
- For proving a sentence is valid, then the negation is unsatisfiable...
Expert Solution & Answer
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Check out a sample textbook solutionStudents have asked these similar questions
Complete the truth table for the implication. You must submit a complete TT. (A ∧ ~B) → C.
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(A ∧ ~B) → C
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T
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F
T
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F
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Are you sure that the negation of the premise is ∃x(Px ∧ ¬∀yPy)? Would it not just be ¬∀x(Px ∧ ¬∀yPy)?
Q1 Show that the argument form with
premises (p A t) →(r V s), q → (u ^ t), u
→p, and ¬s and conclusion q→r is
valid by using rules of inference from
Table 1.
Q2 For each of these arguments, explain
which rules of inference are used for
each step.
a) "Linda, a student in this class, owns a
red convertible. Everyone who owns a
red convertible has gotten at least one
speeding ticket. Therefore, someone in
this class has gotten a speeding ticket."
b) "Each of five roommates, Melissa,
Aaron, Ralph,Veneesha, and Keeshawn,
has taken a course in
discrete mathematics. Every student
who has taken a course in discrete
mathematics can take a course in
algorithms. Therefore, all five
roommates can take a course in
algorithms next year."
c) "All movies produced by John Sayles
are wonderful. John Sayles produced a
movie about coal miners. Therefore,
there is a wonderful movie about coal
miners."
d) "There is someone in this class who
has been to France. Everyone who goes
to France visits the Louvre.…
Chapter 7 Solutions
Artificial Intelligence: A Modern Approach
Ch. 7 - Suppose the agent has progressed to the point...Ch. 7 - (Adapted from Barwise and Etchemendy (1993).)...Ch. 7 - Prob. 3ECh. 7 - Which of the following are correct? a. False |=...Ch. 7 - Prob. 5ECh. 7 - Prob. 6ECh. 7 - Prob. 7ECh. 7 - We have defined four binary logical connectives....Ch. 7 - Prob. 9ECh. 7 - Prob. 10E
Ch. 7 - Prob. 11ECh. 7 - Prob. 12ECh. 7 - Prob. 13ECh. 7 - Prob. 14ECh. 7 - Prob. 15ECh. 7 - Prob. 16ECh. 7 - Prob. 17ECh. 7 - Prob. 18ECh. 7 - A sentence is in disjunctive normal form (DNF) if...Ch. 7 - Prob. 20ECh. 7 - Prob. 21ECh. 7 - Prob. 23ECh. 7 - Prob. 24ECh. 7 - Prob. 25ECh. 7 - Prob. 26ECh. 7 - Prob. 27E
Knowledge Booster
Similar questions
- Question 3 VX(P(X) v Q(X))→ (VXP(X) V VXQ(X)) The above expression follows from the valid argument forms of logic and the rules for quantifiers. True False Question 4 Give an interpretation (in words) of the predicates in the previous question that shows you understand why your answer is correct.arrow_forwardPlease help explain this question into detail. I want to understand the procedure. Thank you. For each of the following, check if it is a tautology, contradiction, or neither.(a) ¬(p ∧ q) → (p → ¬q)(b) ¬(p ∧ q) → (¬p ∧ ¬q)(c) ¬(p ∧ q) ∧ p ∧ qarrow_forward1. Construct a detailed truth table for the following propositional statement: (((P ∨Q)∧¬P)→Q) Is it a tautology? 2. Construct a detailed truth table for the following propositional statement: (P ∧ Q) → R Is it a tautology?arrow_forward
- Classify the following sentence as True or False: Reichenbach’s causation definition solves the problem of Alternative Explanations.arrow_forwardUsing the same defifinitions as in the previous question, translate each of the following predicate logic statements into English. Try to make your English translations as natural sounding as possible. These statements appear long, but try to break them into smaller cohesive pieces. ∀w ∈ K,(T(w) ∧ W(w)) →(∃p ∈ K, P(p) ∧ S(p) ∧ I(p) ∧ A(p, w)). 2. ∃k ∈ K, ∃r ∈ K, k =r ∧ T(k) ∧ T(r) ∧ F(k, r) ∧ F(r, k) ∧ (∀z ∈ K, Z(z) → A(z, k)). 3. ∃t ∈ K, T(t) ∧ W(t) ∧ ∃x ∈ K, ∃y ∈ K, x = y ∧ Z(x) ∧ W(x) ∧ Z(y) ∧ W(y) ∧ A(t, x) ∧ A(t, y) ∧ (∀w ∈ K,(x = w ∧ y = w ∧ Z(w) ∧ W(w)) → ∼ A(t, w))arrow_forwardProposition (Distributive Law): For expressions p1, P2, P3, any word matching the regular expression (P1(P2|P3)) also matches the regular expression ((PıP2)|(P1P3)) Give a proof of the above proposition, or demonstrate that it is false.arrow_forward
- Does the fact ¬Spouse(George,Laura) follow from the facts Jim≠George and Spouse(Jim,Laura)? If so, give a proof; if not, supply additional axioms as needed. What happens if we use Spouse as a unary function symbol instead of a binary predicate?arrow_forward5. Check whether p →r is a valid conclusion from the following premises: p → q V ¬r, q → p^r 6. Using indirect method, prove that q is a valid conclusion from the following premises: p → q,r → q, s → (p V r) 7. Using conditional proof, show that ¬q → s can be derived from the following premises: p → (q → r),¬r V p, qarrow_forwardThis question was on a homework assignment which I could not complete before the deadline. Show that whether the following propositions is a tautology, satisfiable but not a tautology, or a contradiction. If it is a tautology or a contradiction, please give the proof. If it is satisfiable, please give a true assignment and a false assignment. (A ∨ B ∨ ¬C) ∧ (A ∨ ¬B ∨ D) ∧ (A ∨ ¬C ∨ ¬D) ∧ (¬A ∨ ¬B ∨ ¬D) ∧ (A ∨ B ∨ ¬D)arrow_forward
- Prove the following logical implication (A ⇒ B) ⇒ ((C ⇒ A) ⇒ (C ⇒ B)) without a truthtable.arrow_forwardConstruct a truth table for the given statement. Identify whether the statement is a tautology. (-q-p) → (q^p) P T T F F q (-q-p) → (q^ p) T F T F ➤ ➤ *** Is the statement a tautology? OA. The statement is not a tautology, since it is false for all combinations of truth values of the components. OB. The statement is not a tautology, since there is at least one combination of truth values for its components where the statement is false. OC. The statement is a tautology, since there is at least one combination of truth values for its components where the statement is true. OD. The statement is a tautology, since it is true for all combinations of truth values of the components.arrow_forwardDetermine whether or not the following statement is a tautology or not and give reasoning. If you need to, you can build a truth table to answer this question. (q→p)∨(∼q→∼p) A. This is a tautology because it is always true for all truth values of p and q. B. This is not a tautology because it is always false for all truth values of p and q. C. This is a tautology because it is not always false for all values of p and q. D. This is not a tautology becasue it is not always true for all truth values of p and q.arrow_forward
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