Artificial Intelligence: A Modern Approach
3rd Edition
ISBN: 9780136042594
Author: Stuart Russell, Peter Norvig
Publisher: Prentice Hall
expand_more
expand_more
format_list_bulleted
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
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Tick each statement that is valid.
Select one or more:
a. ¬(C → (D → C)) is unsatisfiable.
b.
AV-Bv CV DVE V F V¬A is a tautology
(¬B → B) → (B → C) is a tautology
AV (B → C) v (D → E) v ¬A is a taulogy.
e. (U → U) → U is a tautology.
P→ Q and¬P ^ Q are logically equivalent
g. TVW V-T is a tautology
C.
f.
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
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_forwardUse 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_forwardModule 6 Journal. Please complete each of the proofs below. The proofs below may use any of the rules of inference or replacement rules given in Chapter 8. (D→C) • (C→ D), E → -(D → C) :. -E F • (G V H), -F V -H : F. G -U → ~B, S → ~B, ~(U • -S), TV B .:. T 3. (QR) V (-Q • -R), N → ~(Q ↔ R), EV N.. E -X → -Y, -Xv -Y, Z →Y: -Z -M V N, -R→-N : M → Rarrow_forward
- Consider 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_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_forwardRewrite 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_forward
- 1. 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_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_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_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_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_forwardFor this question about predicate logic, please note that, even though the 'nonsense' words are only nouns and verbs, you do not need to know the meaning of the words being used in order to answer this question. Consider a universe of discourse that contains (among other things) all gudgeons. If the predicates B(x), C(x), and G(x) represented the assertions x brabbles, x corrades, and x groaks, respectively, then which of the following would be an accurate translation of the following assertion? some gudgeons do not brabble even though they corrade and groak Select one: (-B(=) ^ C(z) ^ G(x)) O none of these options (-B(2) v C(x) V G(=)) O Vz (-B(x) ^ C(2) V G(z) O I (¬B(x) V C(x) ^ G(x)) (-B(2) ^ C(x) V G(=)) (-B(=) ^ C(x) V G(=) O Frarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Database System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill Education
Starting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSON
Digital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSON
C How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSON
Database Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage Learning
Programmable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education
Database System Concepts
Computer Science
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:McGraw-Hill Education
Starting Out with Python (4th Edition)
Computer Science
ISBN:9780134444321
Author:Tony Gaddis
Publisher:PEARSON
Digital Fundamentals (11th Edition)
Computer Science
ISBN:9780132737968
Author:Thomas L. Floyd
Publisher:PEARSON
C How to Program (8th Edition)
Computer Science
ISBN:9780133976892
Author:Paul J. Deitel, Harvey Deitel
Publisher:PEARSON
Database Systems: Design, Implementation, & Manag...
Computer Science
ISBN:9781337627900
Author:Carlos Coronel, Steven Morris
Publisher:Cengage Learning
Programmable Logic Controllers
Computer Science
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education