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Chapter 6: Problem 11

Reduce the following Boolean expressions by taking complements: (a) \(\sqrt{[\overline{(\overline{p \cdot q}) \cdot p}][\overline{(\overline{p \cdot q}) \cdot q}]}\) (b) \(\overline{(\overline{p+q+\bar{r}}) \cdot(\overline{p \cdot q}+\overline{r \cdot s})+\overline{q \cdot r \cdot s}}\) (c) \(\overline{(\overline{p \cdot q \cdot r+q \cdot \bar{r} \cdot s})+(\overline{\overline{q \cdot r \cdot s}+\bar{q} \cdot \bar{r} \cdot \bar{s}+q \cdot r \cdot \bar{s}})}\)

Short Answer

Expert verified
(a) Simplifies to p 鈥� q. (b) Simplifies to 1 (tautology). (c) Simplifies to 0 (contradiction).

Step by step solution

01

Simplify the Inside of the Root for (a)

Let's begin by simplifying the expression inside the square root: \[\sqrt{[\overline{(\overline{p \cdot q}) \cdot p}][\overline{(\overline{p \cdot q}) \cdot q}]}\]Firstly, simplify \((\overline{p \cdot q})\) which is equivalent to \[(\bar{p} + \bar{q})\]. Thus, the expression inside becomes:\[[\overline{(\overline{p} + \overline{q}) \cdot p}][\overline{(\overline{p} + \overline{q}) \cdot q}]\]Now, apply De Morgan's Theorems to simplify each term:- For \((\overline{p} + \overline{q}) \cdot p\), it simplifies to \(\bar{p} + \bar{q} + \bar{p}\) which simplifies to \(\bar{p} + \bar{q}\). Thus, \(\overline{\bar{p} + \bar{q}}\) simplifies to \((p \cdot q)\).- Similarly, \([\overline{(\overline{p} + \overline{q}) \cdot q}]\) also simplifies to \((p \cdot q)\).Therefore, the simplified form inside the square root is:\[\sqrt{(p \cdot q) \cdot (p \cdot q)} = \sqrt{p\cdot q}\].
02

Final Simplification for (a)

Knowing the root function in Boolean simplification directly returns the argument when it is a perfect square, we have:\[\sqrt{p\cdot q} = 1\cdot(p \cdot q)=p\cdot q\]Thus, our final simplified answer is \(p \cdot q\).
03

Simplify Expression for (b)

Now, let's work on simplifying the expression:\[\overline{(\overline{p+q+\bar{r}}) \cdot(\overline{p \cdot q}+\overline{r \cdot s})+\overline{q \cdot r \cdot s}}\]Applying De Morgan鈥檚 Theorem:- The inside of \(\overline{p+q+\bar{r}})\) is \(\bar{p}\cdot\bar{q}\cdot(r)\). - The term \((\overline{p \cdot q}+\overline{r \cdot s})\) simplifies to \((\bar{p}+\bar{q})\cdot(r+s)\).- Lastly, the inverted term \(\overline{q \cdot r \cdot s}\) simplifies to \((\bar{q}+\bar{r}+\bar{s})\).The overall expression becomes:\[[(\bar{p}\cdot\bar{q}\cdot r)\cdot(\bar{p}+\bar{q})\cdot(r+s)] + (\bar{q}+\bar{r}+\bar{s})\]This further simplifies to '1' which indicates a tautology in Boolean algebra after breaking into individual minterms.
04

Simplify Expression for (c)

We need to simplify:\[\overline{(\overline{p \cdot q \cdot r+q \cdot \bar{r} \cdot s})+(\overline{\overline{q \cdot r \cdot s}+\bar{q} \cdot \bar{r} \cdot \bar{s}+q \cdot r \cdot \bar{s}})}\]Using De Morgan鈥檚 Laws:- \(\overline{p \cdot q \cdot r + q \cdot \bar{r} \cdot s}\) translates to \((\bar{p}+\bar{q}+\bar{r})\cdot(\bar{q}+r+\bar{s})\).- Similarly, \(\overline{\overline{q \cdot r \cdot s}+\bar{q}\cdot\bar{r}\cdot\bar{s}+q\cdot r\cdot\bar{s}}\) becomes a multivariate simplification, ultimately resolves into '0', indicating a contradiction function. Thus:\[[(\bar{p}+\bar{q}+\bar{r})\cdot(\bar{q}+r+\bar{s})] + 0 = 0\]Thus the simplified result of the expression after considering all internal possibilities is 0.

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Key Concepts

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

De Morgan's Theorem
De Morgan's Theorem is a fundamental principle in Boolean algebra that deals with the complement of a compound expression. It provides a way to simplify expressions involving NOT operations and is particularly useful in the process of logic circuit simplification. De Morgan's Theorem primarily consists of two rules that relate conjunction (AND, \(\cdot\)) and disjunction (OR, \(+\)) through negation:
  • The complement of a conjunction is the disjunction of the complements: \(\overline{A \cdot B} = \bar{A} + \bar{B}\).
  • The complement of a disjunction is the conjunction of the complements: \(\overline{A + B} = \bar{A} \cdot \bar{B}\).
Applying these rules allows for the expression of a complex complemented function in terms of simpler, uncomplemented functions. This makes it easier to work with, especially when designing circuits or simplifying Boolean expressions. For example, in problem (a) of the exercise, De Morgan's Theorem was applied to transform complements of products and sums into simpler expressions. Understanding this theorem is crucial when learning to manipulate and simplify Boolean expressions effectively.
Boolean Simplification
Boolean simplification is the process of reducing Boolean expressions to their simplest form. The goal is to minimize the number of terms and operations without changing the expression's functionality. This is particularly important in designing logic circuits as it can reduce the cost and complexity.

There are various techniques to achieve simplification, one of which is using laws and theorems of Boolean algebra, including:
  • Identity laws
  • Null laws
  • Idempotent laws
  • Complement laws
  • Associative, Distributive, and Absorption laws
In the step-by-step solution given, simplification was applied progressively in each exercise. For instance, part (b) required transforming a combination of terms into simpler segments utilizing De Morgan's Theorem and other Boolean identities. The solution showed how breaking down a problem and applying appropriate laws leads to a dramatically reduced expression, reaching the final form of '1', which represents a Boolean tautology. Mastery in simplification not only helps solve textbook problems but also empowers students to design efficient logical systems.
Boolean Expressions
Boolean expressions are elements of Boolean algebra representing logical propositions using variables and logical operators. These expressions are used to depict relationships and operations among binary variables, typically in digital circuits and computational logic.

A Boolean variable can have one of two values: true (1) or false (0). Common operators include AND (denoted by \(\cdot\)), OR (denoted by \(+\)), and NOT (denoted by \( \bar{} \)). Boolean expressions can get quite complex, involving nested operations and multiple variables.
  • Expression examples include simple forms like \(A \cdot B\) or \(A + \bar{B}\), and more complex configurations with multi-level operations.
  • They are foundational in creating logic gates and circuits used in digital electronics.
Exploring exercises like the ones provided involves transforming and simplifying these expressions. In question (c), the expression incorporated multiple layers and required meticulous application of Boolean principles to reach the simplest form, zero (0). This outcome represents a contradiction, where no combination of input values satisfy the expression. Gaining skill in working with these expressions is vital for students aiming to delve into fields such as computer science, electrical engineering, or information technology, where logic design is a key component.

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