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 17E
a.
Explanation of Solution
Resolution
- The negated goal is ￢G.
- The last two clauses is resolved to produce ￢C and ￢D...
b.
Explanation of Solution
Clauses
- First, each 2-CNF clause has two places to put literals...
c.
Explanation of Solution
Propositional resolution
- Resolving two 2-CNF clauses cannot increase the clause size...
d.
Explanation of Solution
Argument
- First, the number of 3-CNF clauses is O(n3), so for nonpolynomial complexity on the basis of the number of different clauses...
Expert Solution & Answer
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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.
In Boolean logic, a sentence is in 3-Conjunctive Normal Form (abbreviated 3-CNF) if it of the form
C¡ ^ C2 ^ C3 ^ .…. A Cm, where S is the conjunction of m clauses, with each clause being a disjunction of
S
three literals.
For example, the following sentence is in 3-CNF, with m=4 clauses.
(A v B v č) ^ (A v c v D)^ (B v
vc v D)A (a v E v D)
S =
A V B V C
A V C
D) A
B V
В
For example, the first clause is equivalent to this in English: "either A or B or Not C". As long as one of these three
literals is TRUE then this clause evaluates to TRUE. If none of these literals is TRUE (i.e., A=FALSE, B=FALSE,
C=TRUE), then this clause evaluates to FALSE.
We say that S is satisfiable if there exists an assignment of TRUE/FALSE values to each variable (A, B, C, D) so that the
entire sentence S evaluates to TRUE. What this means is that each of the m clauses must evaluate to TRUE.
The above sentence indeed is satisfiable; one possible assignment is A=TRUE, B=FALSE, C=FALSE, D=TRUE.
The first two…
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
- if p and q are logical variables, which of the following is a tautology (i.e., always correct irrespective of specific value of variables) Select one: a. p → (q ∧ p) b. p ∨ (q → q) c. (p ∨ q) → q d. p ∨ (p → q)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_forwardIn Boolean logic, a sentence is in 3-Conjunctive Normal Form (abbreviated 3-CNF) if it of the form C1 ^ C2 ^ C3 ^ .…. A Cm, where S is the conjunction of m clauses, with each clause being a disjunction of S three literals. We say that S is satisfiable if there exists an assignment of TRUE/FALSE values to each variable so that the entire sentence S evaluates to TRUE. For example, in the last class, we showed that the following 3-CNF sentence, with m=8 clauses, is not satisfiable. D) ^ (B v č v D) ^ (c vĀ v D) ^ (A (A v (Bv č v D) A (c v Ā v D) a (A v B V C) ^ (a v B v C) A V v B v D) S = AVВV A V B Let 3-SAT be the decision problem that inputs a 3-CNF sentence S, and outputs YES if S is satisfiable, and outputs NO if S is not satisfiable. Consider these four statements. A. 3-SAT is in P, but is almost surely not in NP B. 3-SAT is in NP, but is almost surely not in P C. 3-SAT is in P and is also in NP D. 3-SAT is neither in P nor in NP Determine which of these four statements is correct.…arrow_forward
- Let p and q be two propositions. Consider the following two statements in prepositional logic. S₁: (-p^(pv q)) → p S₂: q→(-p^ (pvq)) Which of them is tautologyarrow_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_forwardI need help to answer Question no. 30 prove that the below WFF is a valid argument. (∀x)(∀y)[(P(x) Λ S(x,y)) →Q(y)] Λ (∃x)B(x) Λ (∀x)(B(x) → P(x)) Λ (∀x)(∃y)S(x,y) → (∃x)Q(x) Chapter 1.4 - Predicate Logic Text Book: Discrete Mathematics and its application 7th edition. Author: Judith L Gersting.arrow_forward
- Exercise 1.4.1: Proving tautologies and contradictions. About Show whether each logical expression is a tautology, contradiction or neither. (a) (p ∨ q) ∨ (q → p) (b) (p → q) ↔ (p ∧ ¬q) (c) (p → q) ↔ p (d) (p → q) ∨ p (e) (¬p ∨ q) ↔ (p ∧ ¬q) (f) (¬p ∨ q) ↔ (¬p ∧ q)arrow_forwardPart 1: Proof-Theoretic Concepts Show that each of the following pairs of sentences are provably equivalent in SL 1. P → R, ¬R → ¬Parrow_forward8. Simplify the given expression so that negations appear only within individual predicates (that is, so that no negation is outside an expression involving logical connectives). -[(p → -q)A¬(-¬q ^p)) → r] =arrow_forward
- use propositional logic to see if the argument is valid (A ∧ B) ∧ (B → A’) → (C ∧ B’) A ∧ B Hypotheses B → A’ Hypotheses Chart to go off of attached belowarrow_forwardProve the predicate logic statement is valid: (∀ x)(P(x)) ∧ (∀ x)(Q(x)) → P(a) ∧ Q(b)arrow_forward1: Prove the validity of the following sequents : a. (cAn) → t, hA-s, hA-(s V c) pE (n A¬t) → p b. (s – p) V (t → q) E (s → q) V (t → p) C. vx (P(x) → (Q(x) V R(x))), – 3x (P(x) A R(x)) Fvx (P(x) → Q(x))arrow_forward
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