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
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))
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
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
Similar questions
- Use De Morgan’s Laws, and any other logical equivalence facts you know to simplify the followingstatements. Show all your steps. Your final statements should have negations only appear directly nextto the sentence variables (P, Q, etc.), and no double negations. It would be a good idea to use onlyconjunctions, disjunctions, and negations.(a) ¬((¬P ∧ Q) ∨ ¬(R ∨ ¬Q)).(b) ¬((¬P → ¬Q) ∧ (¬Q → R)) (careful with the implications).(c) For both parts above, verify your answers are correct using truth tables. That is, use a truth tableto check that the given statement and your proposed simplification are actually logically equivalent.arrow_forward8. 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_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_forward
- Rewrite the following statement so that negations appear only within predicates (that is, no negation is outside a quantifier or an expression involving logical connectives) ¬∀x ((∀y∃z P(x, y, z)) ⊕ (∀z∀y (R(x, y, z) ∨ S(z, y)))).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_forward3. Write a sentence that is true in an interpretation A if, and only if, the domain of A contains exactly two elements. 4. Write a sentence that is true in an interpretation A if, and only if, the domain of 2A contains exactly three elements.arrow_forward
- a) 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_forwardExercise 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_forwardProposition (Distributive Law): For expressions P1, P2, P3, any word matching the regular expression (P1 (P2|P3)) also matches the regular expression ((P1P2) (P1P3)) Give a proof of the above proposition, or demonstrate that it is false.arrow_forward
- Let P(x), Q(x) and R(x) be the statements "is a duck", "is annoying" and "is a dancer" respectively. Express each of these statements using quantifiers, logical connectives, and P(x), Q(x) and R(x). a) All ducks are annoying. b) Some dancers are not annoying.arrow_forwardConstruct a truth table for (p ∨ ¬ q) ∨ (¬ p ∧ q) Use the truth table that you constructed in part 1 to determine the truth value of (p ∨¬q) ∨ (¬ p ∧ q), given that p is true and q is false. Determine whether the given statement is a tautology, contradiction, or contingency. p V (~p V q) ~ (p ∧ q) ~p V ~qarrow_forwardPart 1: Proof-Theoretic Concepts Show that each of the following pairs of sentences are provably equivalent in SL 1. P → R, ¬R → ¬Parrow_forward
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