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Chapter 1: Problem 47

Construct a combinatorial circuit using inverters, OR gates, and AND gates that produces the output \(((\neg p \vee \neg r) \wedge \neg q) \vee(\neg p \wedge(q \vee r))\) from input bits \(p, q\) and \(r .\)

Short Answer

Expert verified
The circuit uses inverters for \( p, q, r \), OR gates to combine terms, and AND gates to form intermediate results, finalizing with an OR gate to derive the output.

Step by step solution

01

Identify the Subexpressions

Break down the expression \(((\eg p \vee \eg r) \wedge \eg q) \vee(\eg p \wedge(q \vee r))\) into smaller subexpressions to work with. 1. \(a = \eg p\)2. \(b = \eg r\)3. \(c = \eg q\)
02

Construct Inverters

Use inverters to get \(\eg p\), \(\eg q\), and \(\eg r\) from input bits \(p\), \(q\), and \(r\). 1. Inverter for \(p\) gives \(\eg p\). 2. Inverter for \(q\) gives \(\eg q\). 3. Inverter for \(r\) gives \eg r\.
03

Combine Inverters with OR Gates

Combine the outputs of inverters using OR gates: 1. OR gate for \(\eg p \vee \eg r\).2. OR gate for \((q \vee r)\). 1. Combine \(\eg p\) and \(\eg r\) using OR gate: \(d = \eg p \vee \eg r\). 2. Combine \(q\) and \(r\) using OR gate: \(e = q \vee r\).
04

Construct AND Gates

Use AND gates to form intermediate products: 1. Combine \(d\) and \(\eg q\): \(f = (\eg p \vee \eg r) \wedge \eg q\). 2. Combine \(\eg p\) and \(e\): \(g = \eg p \wedge (q \vee r)\).
05

Final OR Gate

Combine the outputs of the two AND gates using an OR gate to get the final output: \( OP = f \vee g = ((\eg p \vee \eg r) \wedge \eg q) \vee (\eg p \wedge (q \vee r)) \)

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Key Concepts

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

logic gates
Logic gates are the building blocks of digital circuits. They perform basic logical functions that are essential for creating complex electronics systems.
There are several types of logic gates, each performing a different logical operation. The most common ones are:
  • AND gate
  • OR gate
  • NOT gate (Inverter)

These gates take one or more binary inputs (0 or 1) and produce a single binary output. For example, an AND gate outputs 1 only when all its inputs are 1. Understanding how these gates work is crucial for constructing combinatorial circuits.

boolean algebra
Boolean algebra is a branch of mathematics that deals with true or false values, commonly represented as 1 and 0. It provides the foundation for designing and analyzing digital circuits.
In Boolean algebra, there are several basic operations:
  • AND (denoted as \( \wedge \))
  • OR (denoted as \( \vee \))
  • NOT (denoted as \( eg \))

Boolean expressions can be simplified to minimize the number of logic gates required in a circuit. The expression \(((eg p \vee eg r) \wedge eg q) \vee (eg p \wedge (q \vee r))\), for example, uses AND, OR, and NOT operations to get the desired output.

digital circuits
Digital circuits are electronic circuits that operate using digital signals. These circuits are designed using logic gates, and they perform a variety of functions including computation, data processing, and control operations.
Digital circuits can be divided into two main types:
  • Combinatorial Circuits: where the output only depends on the current inputs.
  • Sequential Circuits: where the output depends on the current inputs and the past sequence of inputs.

In the given exercise, a combinatorial circuit is designed to produce a specific output based on the input bits p, q, and r.

inverters
Inverters, also known as NOT gates, are a type of logic gate that outputs the opposite value of the input. If the input is 1, the output is 0, and vice versa.
In the given exercise, inverters are used to obtain \( eg p \), \( eg q \), and \( eg r \) from the input bits p, q, and r. This is the crucial first step in constructing the desired complex Boolean expression.

OR gates
OR gates are logic gates that output 1 if any of the inputs are 1. The output is only 0 when all inputs are 0.
In the step-by-step solution, OR gates are used to combine the outputs of the inverters:
  • To get \( eg p \vee eg r \)
  • To get \( q \vee r \)

These intermediate results are then used in subsequent steps to form the desired final output of the combinatorial circuit.
AND gates
AND gates are logic gates that output 1 only if all the inputs are 1. The output is 0 if any of the inputs are 0.
In the exercise, AND gates are used to combine the results from the OR gates and inverters:
  • To get \( (eg p \vee eg r) \wedge eg q \)
  • To get \( eg p \wedge (q \vee r) \)

Finally, these intermediate results are combined using another OR gate to achieve the complex Boolean expression required by the exercise.

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Most popular questions from this chapter

Determine whether these are valid arguments. a) If \(x\) is a positive real number, then \(x^{2}\) is a positive real number. Therefore, if \(a^{2}\) is positive, where \(a\) is a real number, then \(a\) is a positive real number. b) If \(x^{2} \neq 0,\) where \(x\) is a real number, then \(x \neq 0 .\) Let \(a\) be a real number with \(a^{2} \neq 0 ;\) then \(a \neq 0\)

Justify the rule of universal transitivity, which states that if \(\forall x(P(x) \rightarrow Q(x))\) and \(\forall x(Q(x) \rightarrow R(x))\) are true, then \(\forall x(P(x) \rightarrow R(x))\) is true, where the domains of all quantifiers are the same.

Use quantifiers to express the associative law for multiplication of real numbers.

Exercises \(61-64\) are based on questions found in the book Symbolic Logic by Lewis Carroll. Let P(x), Q(x), and R(x) be the statements 鈥渪 is a professor,鈥� 鈥渪 is ignorant,鈥� and 鈥渪 is vain,鈥� respectively. Express each of these statements using quantifiers; logical connectives; and P(x), Q(x), and R(x), where the domain consists of all people. $$ \begin{array}{l}{\text { a) No professors are ignorant. }} \\ {\text { b) All ignorant people are vain. }} \\ {\text { c) No professors are vain. }} \\\ {\text { d) Does (c) follow from (a) and (b)? }}\end{array} $$

A statement is in prenex normal form (PNF) if and only if it is of the form $$ Q_{1} x_{1} Q_{2} x_{2} \cdots Q_{k} x_{k} P\left(x_{1}, x_{2}, \ldots, x_{k}\right) $$ where each \(Q_{i}, i=1,2, \ldots, k,\) is either the existential quantifier or the universal quantifier, and \(P\left(x_{1}, \ldots, x_{k}\right)\) is a predicate involving no quantifiers. For example, \(\exists x \forall y(P(x, y) \wedge Q(y))\) is in prenex normal form, whereas \(\exists x P(x) \vee \forall x Q(x)\) is not (because the quantifiers do not all occur first). Every statement formed from propositional variables, predicates, \(\mathbf{T},\) and \(\mathbf{F}\) using logical connectives and quantifiers is equivalent to a statement in prenex normal form. Exercise 51 asks for a proof of this fact. Show how to transform an arbitrary statement to a statement in prenex normal form that is equivalent to the given statement. (Note: A formal solution of this exercise requires use of structural induction, covered in Section \(5.3 . )\)
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