/*! 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 Use the De Morgan laws to negate... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the De Morgan laws to negate the function $$ f=(p+q) \cdot(\bar{r} \cdot s) \cdot(q+\bar{t}) $$

Short Answer

Expert verified
The negated function is \((\neg p \cdot \neg q) + (r + \bar{s}) + (\bar{q} \cdot t)\).

Step by step solution

01

Apply De Morgan's Law to the First Term

Using De Morgan's Law: \(eg (A + B) = eg A \cdot eg B\), negate the first term \((p + q)\). This becomes \(eg p \cdot eg q\).
02

Apply De Morgan's Law to the Second Term

Using De Morgan's Law: \(eg (A \cdot B) = eg A + eg B\), negate the second term \((\bar{r} \cdot s)\). This becomes \(r + \bar{s}\).
03

Apply De Morgan's Law to the Third Term

Using De Morgan's Law: \(eg (A + B) = eg A \cdot eg B\), negate the third term \((q + \bar{t})\). This becomes \(\bar{q} \cdot t\).
04

Combine Negated Terms using De Morgan's Law for the Whole Expression

After negating all individual terms, De Morgan's Law \(eg (A \cdot B \cdot C) = eg A + eg B + eg C\) is used to combine them. This results in \((eg p \cdot eg q) + (r + \bar{s}) + (\bar{q} \cdot t)\).

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Key Concepts

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

Boolean Algebra
Boolean algebra is a mathematical framework that deals with binary variables and logical operations.It uses the binary values of 0 and 1 to represent the logic of true and false.This algebra is named after George Boole, who first defined it in his work on logic.
The fundamental operations of Boolean algebra include the logical AND (denoted by \( \cdot \)), logical OR (denoted by \( + \)), and logical NOT (or negation, denoted by an overline, e.g., \( \bar{A} \)).
These operations are similar to arithmetic but follow different rules.
  • AND operation returns true if both operands are true.
  • OR operation returns true if at least one operand is true.
  • NOT operation returns the opposite value of the operand.
Boolean algebra is crucial for designing and understanding digital circuits as well as expressing logical statements efficiently.De Morgan's Laws offer critical insights into how these logical statements can be manipulated and simplified in Boolean terms.
Logical Negation
Logical negation is a key concept in logic and Boolean algebra, represented by the NOT operation.It involves flipping the truth value of a proposition, meaning it converts a true statement to false, and vice versa.In Boolean algebra, this is shown with an overline or bar over the variable, such as \( \bar{A} \).
Logical negation plays a vital role in forming complementary logical expressions.For example, knowing how to properly negate components in expressions and equations can lead to simplified results.
According to De Morgan's Laws:
  • The negation of an AND operation becomes an OR operation with negated operands: \( \overline{A \cdot B} = \bar{A} + \bar{B} \).
  • The negation of an OR operation becomes an AND operation with negated operands: \( \overline{A + B} = \bar{A} \cdot \bar{B} \).
These laws show how negation can transform the logical structure of complex expressions.
Propositional Logic
Propositional logic, also known as statement logic or sentential logic, is an area of logic that deals with propositions.A proposition is a declarative statement that is either true or false.In Boolean algebra, these propositions are often represented by variables such as \( p, q, \) and \( r \).
One of the essential operations in propositional logic is constructing compound statements using logical connectives.These include AND, OR, and NOT, akin to the operations in Boolean algebra.De Morgan's Laws play an important role here by allowing the restructuring of compound statements through negation.
For instance:
  • The negation of an OR compound becomes an AND compound with negated components: \( eg(p \lor q) = eg p \land eg q \).
  • The negation of an AND compound becomes an OR compound with negated components: \( eg(p \land q) = eg p \lor eg q \).
Understanding these fundamental tools equips students to analyze and transform logical expressions effectively.

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Most popular questions from this chapter

Let \(A=\\{n \in \mathbb{N}, n \leqslant 11\\}\) $$ \begin{aligned} &B=\\{n \in \mathbb{N}, n \text { is even and } n \leqslant 20 \\ &C=\\{10,11,12,13,14,15,17,20 \end{aligned} $$ Write down the sets (a) \(A \cap B\) (b) \(A \cap B \cap C\) (c) \(A \cup(B \cap C)\) and verify that \((A \cup B) \cap(A \cup C)=A \cup(B \cap C)\)

Determine which of the following are propositions and which are not. For those that are, determine their truth values. (a) \(x+y=y+x\) for all \(x, y \in \mathbb{R}\). (b) \(A B=B A\), where \(A\) and \(B\) are square matrices. (c) Academics are absent-minded. (d) 1 think that the world is flat. (e) Go fetch a policeman. (f) Every even integer greater than 4 is the sum of two prime numbers. (This is Goldbach's conjecture.)

Defining the difference \(A-B\) between two sets \(A\) and \(B\) belonging to the same universal set \(U\) to be the set of elements of \(A\) that are not elements of \(B\), that is \(A-B=A \cap \bar{B}\), verify the following properties: (a) \(U-A=\bar{A}\) (b) \((A-B) \cup B=A \cup B\) (c) \(C \cap(A-B)=(C \cap A)-(C \cap B)\) (d) \((A \cup B) \cup(B-A)=A \cup B\) Illustrate the identities using Venn diagrams.

Draw switching circuits to establish the truth of the following laws: (a) \(p+p \cdot q=p\) (b) \(p+\bar{p} \cdot q=p+q\). (c) \(p \cdot q+p \cdot r=p \cdot(q+r)\) (d) \((p+q) \cdot(p+r)=p+q \cdot r\) Use these to simplify the expression. $$ s=p \cdot \bar{p}+p \cdot q+\bar{p} \cdot r+q \cdot r $$ so that \(s\) only contains two pairs of products added.

Design a switching circuit containing three independent contacts for a machine so that the machine is turned on when any two, but not three, of the contacts are closed.

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