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 13E
a.
Explanation of Solution
Clause
- P ⇒ Q is equivalent to ￢P ∨Q by implication elimination.
- ￢(P1 ∧· · · ∧ Pm) is equivalent to (￢P1 ∨ · ...
b.
Explanation of Solution
Clause
- A clause can have positive and negative literals.
- Let the negative literals have the form￢P1, . . . ,￢Pm and let the positive literals have the form Q1, . . . ,Qn, where the Pis and Qjs are symbols.
- Then the clause can be written as (￢P1 ∨· &#x...
c.
Explanation of Solution
Full resolution rule
- For atoms pi, qi, ri, si, then pj = qk:
- Then the full resolution rule is
p1 ∧ . . . pj . . . ∧ pn1 ⇒ r1 ∨ . . . rn2
s1 ∧ ...
Expert Solution & Answer
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Students have asked these similar questions
1
Let . Let p and q be the propositions “The election is decided” and “The votes have been counted,” respectively. Express each of these compound propositions as an English sentence. A. ¬p → ¬qB. p ∧ qC. ¬p ∨ qD. ¬p ∧ (p ∨ ¬q)E. p ↔ ¬qF. ¬p → ¬q
2.
For each of the following statements, push all negations (¬) as far as possible so
that no negation is to the left of a quantifier (in other words, the negation is immediately to the left
of a predicate). Show your work.
(az(A(x, z) ^ B(y, z))
(b) y(xA(x, y) → EzB(z, y))
(c)¬\x((³yA(x, y) → ]yB(x, y)) → ¬³yС(x, y))
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.
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
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- 8. Let the universe of discourse be U = {−2,−1, 0, 1, 2}. Compute the negations for each of thefollowing statements, via transforming the negation connectives inward, such that the negationsymbols immediately precede predicates. Thereafter, determine whether each of the negatedstatement is true or false. Note that x and y range over U.(a) ∀x∃y(x + y = 1).(b) ∃x∀y(x + y = −y).(c) ∀x∃y(xy ≥ y).(d) ∃x∀y(x ≤ y).arrow_forwardConsider the following assertions about the sets A, B and C. Write them down in the language of predicate logic. Use only the constructions of predicate logic (∀, ∃, ¬, ⇒, ∧, ∨) and the elementof symbol (∈). Do not use derived notions (∩, ∪, =, etc.). Hint “A is a subset of B” can be formalized as ∀x. x ∈ A =⇒ x ∈ B. (i) The sets A and B are equal. (ii) Every element of A is in the set B or the set C. (iii) If A is disjoint from B then B and C overlap.arrow_forwarda) Let Γ be a set of sentences of L1 (the language ofpropositional logic) and let φ be a sentence of L1.Consider:If the argument from Γ to φ is not valid, theargument from Γ to ¬φ is valid.Is this claim correct? Why or why not?arrow_forward
- Exercise 1 Let B be a set of Boolean variables and P be a propositional logic formula over B. If P always evaluates to 1, no matter than assignments to each b ∈ B, then P is called a tautology. Formulate this definition as an expression in first order predicate logic.arrow_forward1. Teachers in the Middle Ages supposedly tested the real-time propositional logic ability of a student via a technique known as an obligato game. In an obligato game, a number of rounds is set and in each round the teacher gives the student successive assertions that the student must either accept or reject as they are given. When the student accepts an assertion, it is added as a commitment; when the student rejects an assertion its negation is added as a commitment. The student passes the test if the consistency of all commitments is maintained throughout the test. a.) Suppose that in a three-round obligato game, the teacher first gives the student the proposition p → q, then the proposition ¬(p ∨ r) ∨ q, and finally the proposition q. For which of the eight possible sequences of three answers will the student pass the test? b.) Explain why every obligato game has a winning strategy.arrow_forwardForm the negation of the following statements. Then apply De Morgan's law and/or conditional law, when applicable. Negation should appear only within predicates, i.e., no negation should be outside a quantifier or an expression involving logical connectives. Show all steps. a) Vx (P(x) A R(x)) b) Vy³z(-P(y)→→ Q(z)) c) 3x (P(x) V (VZ (R(z) →→Q(z))))arrow_forward
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