/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Which of the following DNF formu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the following DNF formulas are satisfiable? If the formula is satisfiable, give an interpretation that satisfies it. If it is not satisfiable, explain why not. (a) \((a \wedge b \wedge c) \vee(c \wedge \neg c \wedge b)\) (b) \((a \wedge b \wedge c \wedge d \wedge \neg b) \vee(c \wedge d \wedge \neg c \wedge e \wedge f)\) (c) \((a \wedge b \wedge c) \vee(\neg a \wedge \neg b \wedge \neg c)\)

Short Answer

Expert verified
(a) Satisfiable: \(a = b = c = \text{true}\). (b) Not satisfiable. (c) Satisfiable: \(a = b = c = \text{true}\) or all false.

Step by step solution

01

Analyze Formula (a)

The formula is \((a \wedge b \wedge c) \vee(c \wedge eg c \wedge b)\). This is a disjunction of two clauses. The clause \(a \wedge b \wedge c\) involves three variables, and it is true if all variables are true. The clause \(c \wedge eg c \wedge b\) contains a contradiction because of \(c \wedge eg c\), making this part always false.
02

Determine Satisfiability of Formula (a)

Since \(c \wedge eg c\) is always false, we only need to check the clause \(a \wedge b \wedge c\). The formula is satisfiable because if you assign \(a = \text{true}, b = \text{true}, c = \text{true}\), the entire formula becomes true.
03

Analyze Formula (b)

The formula is \((a \wedge b \wedge c \wedge d \wedge eg b) \vee(c \wedge d \wedge eg c \wedge e \wedge f)\). Both sub-clauses contain contradictions: the first clause has \(b \wedge eg b\), and the second has \(c \wedge eg c\), making both clauses always false.
04

Determine Satisfiability of Formula (b)

Because both clauses contain contradictions and are always false, the entire formula is unsatisfiable. There are no assignments of truth values that can make it true.
05

Analyze Formula (c)

The formula is \((a \wedge b \wedge c) \vee(eg a \wedge eg b \wedge eg c)\). The first sub-clause is true when \(a, b,\) and \(c\) are all true. The second sub-clause is true when \(a, b,\) and \(c\) are all false. This means there are scenarios where either clause can be true.
06

Determine Satisfiability of Formula (c)

Since there are assignments in which either sub-clause is true (such as \(a = \text{true}, b = \text{true}, c = \text{true}\) or \(a = \text{false}, b = \text{false}, c = \text{false}\)), the formula is satisfiable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

DNF (Disjunctive Normal Form)
DNF, or Disjunctive Normal Form, is a way of structuring logical expressions. In DNF, a formula is composed of a series of disjunctions (logical ORs) of conjunctions (logical ANDs) of literals. For example, a formula like \((a \wedge b) \vee c\) is in DNF. Here, \((a \wedge b)\) and \(c\) are the conjunctions combined by a disjunction.

This structure gives DNF a particular advantage when analyzing logical statements for satisfiability. Each conjunction, or clause, is considered individually. A DNF formula is satisfiable if there is at least one clause that can be made true by some assignment of truth values to its variables.

When solving problems in logic, especially in satisfiability problems, rewriting a formula in DNF can simplify the process. It allows you to focus on each clause independently. A simple tip for recognizing DNF is to look for clauses linked by \(\vee\) where each clause is a chain of \(\wedge\) operations.
Contradiction in Logic
Understanding contradictions is crucial when working with logical formulas. A contradiction occurs when a logical expression is inherently false, no matter the truth values assigned to its variables. It often arises when a variable and its negation appear in the same conjunction, such as \(c \wedge eg c\). This scenario is always false because a statement and its negation cannot both be true simultaneously.

In the context of DNF, identifying contradictions helps determine the satisfiability of a formula. If a clause contains a contradiction, that clause can never contribute to making the entire formula true. For instance, in the DNF formula \((b \wedge eg b)\), regardless of any truth assignment, this clause is false.

To solve satisfiability problems, pinpoint clauses with contradictions and consider them as irrelevant for achieving a true formula status. Thus, you focus only on the clauses without contradictions.
Truth Assignment
Truth assignment is the process of determining the truth values (true or false) for the variables in a logical formula. It is essential for establishing whether a formula is satisfiable. Essentially, you're assigning 'true' or 'false' to individual variables to evaluate the formula's overall truthfulness.

In DNF formulas, a truth assignment that makes any one clause true makes the entire formula satisfiable. For example, consider the formula \((a \wedge b \wedge c) \vee (eg a \wedge eg b \wedge eg c)\). You can assign truth values such as \(a = \text{true}, b = \text{true}, c = \text{true}\) for the first clause or \(a = \text{false}, b = \text{false}, c = \text{false}\) for the second. Both truth assignments satisfy the formula.

When analyzing logical expressions, look for truth assignments that fulfill at least one clause without contradiction. If such truth assignments exist, the formula is satisfiable. Hence, becoming skilled in truth assignment is a significant step toward understanding logical expressions and their satisfiability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Find the resolvant of \((p \vee q)\) and \((\neg p \vee r)\) on \(p\). (b) Find the resolvant of \((p \vee q \vee r \vee s)\) and \((\neg p \vee \neg q \vee t)\) on \(p\). (c) Find the resolvant of \((p \vee q)\) and \(\neg p\) on \(p\). (d) Find the resolvant of \((p)\) and \((\neg p)\) on \(p\). (e) Which resolvant above from parts (a) through (d) is a tautology? Which is tautologically false?

For the following formulas, let the universe be \(\mathbb{R}\). Translate each of the following sentences into a formula (using quantifiers): (a) There is no largest number. (b) There is no smallest positive number. (c) Between any two distinct numbers, there is a third number not equal to either of them.

The first stage of the method described to "push negations inward" was a method to climinate \(\rightarrow\) 's and \(\leftrightarrow\) 's. Prove that in the method to eliminate them, the process of substituting always stops, Consider, for example, the substitution in the formula $$ (p \leftrightarrow q) \leftrightarrow(r \leftrightarrow s) $$ If the substitution is first performed on the second \(\leftrightarrow\), the resultant formula is $$ ((p \leftrightarrow q) \rightarrow(r \leftrightarrow s)) \wedge((r \leftrightarrow s) \rightarrow(p \leftrightarrow q)) $$ which has more \(\leftrightarrow\) 's to replace than in the original formula! At first sight, one might expect that if the substitutions are made in the wrong order, the process might continue generating more \(\leftrightarrow\) 's at each stage, and the process might continue forever. (Hint: One method is to, instead of just counting the number of \(\leftrightarrow\) symbols, put a weight on each \(\leftrightarrow\) symbol, with the weight of the \(\leftrightarrow\) symbol in \(\psi \leftrightarrow x\) being dependent on the number of \(\leftrightarrow\) 's in \(\psi\) and \(\chi\). If the correct method of calculating weights is used, it can be shown that the total weight of the \(\leftrightarrow\) 's decreases with each substitution.

Simplify the following boolean expressions: (a) \((x \wedge y) \vee(x \wedge \neg y) \vee(\neg x \wedge y) \vee(\neg x \wedge \neg y)\) (b) \((x \wedge y \wedge z) \vee(x \wedge \neg y \wedge z) \vee(\neg x \wedge y \wedge \neg z) \vee(\neg x \wedge \neg y \wedge z)\) (c) \((x \wedge y \wedge \neg z) \vee(x \wedge \neg y \wedge z) \vee(x \wedge \neg y \wedge \neg z)\)

Write three descriptions of the elements of the set 12,5,8,11,14\(\\}\)

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.