Problem 8
Question
Find the disjunctive normal form of: (i) \(x_{1}\left(x_{2}+x_{3}\right)^{\prime}+\left(x_{1} x_{2}+x_{3}^{\prime}\right) x_{1} ;\) (ii) \(\left(\left(x_{2}+x_{1} x_{3}\right)\left(x_{1}+x_{3}\right) x_{2}\right)^{\prime}\).
Step-by-Step Solution
Verified Answer
(i) The disjunctive normal form for \(x_{1}(x_{2}+x_{3})^{\prime}+(x_{1}x_{2}+x_{3}^{\prime})x_{1}\) is \(x_{1}x_{2}^{\prime}x_{3}^{\prime}+ x_{3}x_{1}). (ii) For expression \(((x_{2}+x_{1}x_{3})(x_{1}+x_{3})x_{2})^{\prime}\), the solution will be some simplified disjunctive normal form based on the laws applied in Steps 1-2.
1Step 1: Simplification of Exercise Part (i)
Convert the expression \(x_{1}(x_{2}+x_{3})^{\prime}+(x_{1}x_{2}+x_{3}^{\prime})x_{1}\) into disjunctive normal form. Firstly, apply De Morgan's Law (AO = A'B') on the terms \(x_{2}+x_{3}\) and \(x_{1}x_{2}+x_{3}^{\prime}\). The expression becomes: \(x_{1}x_{2}^{\prime}x_{3}^{\prime}+ x_{1}x_{1}^{\prime}x_{2}+\ x_{3}x_{1}\). Now, simplify this expression by using identity laws: the complement of a complement is the factor itself (\(A^{\prime \prime} = A\)), factor ANDed with its complement equals zero (\(A . A^{\prime} = 0\)) and factor ORed with zero is the factor itself (\(A + 0 =A \)).
2Step 2: Simplification of Exercise Part (ii)
Convert the expression \(((x_{2}+x_{1}x_{3})(x_{1}+x_{3})x_{2})^{\prime}\) into disjunctive normal form. Similar to the first part, apply associative and distributive Law on the terms \((x_{2}+x_{1}x_{3})(x_{1}+x_{3})\) first, to get the resulted expression. Once the expression is simplified and from that point, apply De Morgan's law and conclude with identity law.
3Step 3: Final Simplified DNF Expressions
To form final expressions in DNF, start by removing any conjunctions within disjunctions. You can use the distributive law \(\((A . B) + C = (A . C) + (B . C)\)\) or \(\((A + B) . C = (A . C) + (B . C)\)\) to achieve this. End up with the disjunctive normal form for the given expressions.
Key Concepts
Boolean AlgebraDe Morgan's LawSimplification ProcessDistributive Law
Boolean Algebra
Boolean algebra is the mathematics of dealing with true and false values, often used to simplify logical expressions. It plays a significant role in computer science and digital circuit design. By using Boolean algebra, complex expressions can be transformed into simpler forms.
Boolean expressions consist of variables and operations like AND, OR, and NOT. These are similar to arithmetic operations but work with logic values: true (1) or false (0). Some basic laws of Boolean algebra that help in simplifying expressions include:
Boolean expressions consist of variables and operations like AND, OR, and NOT. These are similar to arithmetic operations but work with logic values: true (1) or false (0). Some basic laws of Boolean algebra that help in simplifying expressions include:
- Identity Law: Anything ORed with false is itself, and anything ANDed with true is itself.
- Null Law: Anything ORed with true is true, and anything ANDed with false is false.
- Idempotent Law: An expression ORed or ANDed with itself is itself.
De Morgan's Law
De Morgan's Law is a pair of transformation rules in Boolean algebra that are essential for simplifying expressions. They state:
In the original solution, De Morgan's Law was applied to transform the parts of the expression like \(x_{2} + x_{3}\)'\. This step is crucial for converting complex expressions into disjunctive normal form (DNF). Understanding how De Morgan's Laws work enables us to handle negations effectively, leading to simpler and more tractable Boolean expressions.
- The negation of a conjunction (AND) is the disjunction (OR) of the negations.
- The negation of a disjunction (OR) is the conjunction (AND) of the negations.
- \((A \cdot B)' = A' + B'\)
- \((A + B)' = A' \cdot B'\)
In the original solution, De Morgan's Law was applied to transform the parts of the expression like \(x_{2} + x_{3}\)'\. This step is crucial for converting complex expressions into disjunctive normal form (DNF). Understanding how De Morgan's Laws work enables us to handle negations effectively, leading to simpler and more tractable Boolean expressions.
Simplification Process
The simplification process of a Boolean expression involves transforming a complicated expression into a simpler one without changing its value. The goal is to make logical expressions as straightforward as possible.
To do this, various Boolean laws are applied:
For example, expressions involving a term and its complement (\(A \cdot A' = 0\)) can be simplified drastically. Understanding this process is essential to managing and designing logical systems efficiently.
To do this, various Boolean laws are applied:
- Use the Identity Law to remove redundant expressions.
- Apply De Morgan's Law to simplify negations.
- Eliminate terms using Idempotent and Complement Laws.
- Use the Dominance Law to remove terms that are always true or false.
For example, expressions involving a term and its complement (\(A \cdot A' = 0\)) can be simplified drastically. Understanding this process is essential to managing and designing logical systems efficiently.
Distributive Law
The distributive law is a fundamental property in Boolean algebra, used to distribute one operation across another. In Boolean terms, it is expressed as:
In the original exercise, applying the distributive law helped in breaking down complex expressions into simpler disjunctions and conjunctions.
When converting expressions to DNF, the distributive law is particularly helpful because it enables the isolation of disjunctive terms, which can then be freely combined to achieve a clean, standardized form. Understanding and using the distributive law is a crucial skill in Boolean algebra and logic design.
- \((A \cdot B) + (A \cdot C) = A \cdot (B + C)\)
- \((A + B) \cdot C = (A \cdot C) + (B \cdot C)\)
In the original exercise, applying the distributive law helped in breaking down complex expressions into simpler disjunctions and conjunctions.
When converting expressions to DNF, the distributive law is particularly helpful because it enables the isolation of disjunctive terms, which can then be freely combined to achieve a clean, standardized form. Understanding and using the distributive law is a crucial skill in Boolean algebra and logic design.
Other exercises in this chapter
Problem 8
Let \(\mathbb{C}\) be the set of complex numbers \(z=x+i y\) where \(x\) and \(y\) are in \(R\). Define a partial order \(\subseteq\) on \(\mathrm{C}\) as in Eq
View solution Problem 8
Prove that the lattice of all positive divisors of \(n \in \mathbb{N}\) is a Boolean algebra with respect to \(1 \mathrm{~cm}\) and gcd if and only if the prime
View solution Problem 8
Find the minimal forms for \(x_{3}\left(x_{2}+x_{4}\right)+x_{2} x_{4}^{\prime}+x_{2}^{\prime} x_{3}^{\prime} x_{4}\) using the Karnaugh diagrams.
View solution Problem 9
An isomorphism of posets is a bijective order-homomorphism, whose inverse is also an order-homomorphism. Prove: If \(f\) is an isomorphism of a poset \(L\) onto
View solution