/*! 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 14 Simplify the following formulas:... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 14

Simplify the following formulas: (a) \((p \Rightarrow \bar{q}) \wedge(\bar{q} \Rightarrow p)\) (b) \(\overline{p \wedge \bar{q}}\) (c) \(p \wedge(\bar{p} \vee q)\)

Short Answer

Expert verified
(a) \(\bar{p} \wedge q \vee \bar{q} \wedge p\), (b) \(\bar{p} \vee q\), (c) \(p \wedge q\).

Step by step solution

01

Simplify Formula (a) - Distribute and Apply Implication

In formula (a), replace the implications: First, replace \(p \Rightarrow \bar{q}\) with \(\bar{p} \vee \bar{q}\), and \(\bar{q} \Rightarrow p\) with \(q \vee p\). The formula becomes:\[(\bar{p} \vee \bar{q}) \wedge (q \vee p)\]This simplifies to the intersection of these two expressions. Now apply distribution.
02

Simplify Formula (a) - Combine Terms

Distribute the terms:\[(\bar{p} \vee \bar{q}) \wedge (q \vee p) = (\bar{p} \wedge q) \vee (\bar{p} \wedge p) \vee (\bar{q} \wedge q) \vee (\bar{q} \wedge p)\]This can be further simplified: since \(\bar{p} \wedge p\) and \(\bar{q} \wedge q\) are contradictions (equal to false), they disappear:\[\bar{p} \wedge q \vee \bar{q} \wedge p\]
03

Simplify Formula (b) - Apply De Morgan's Law

For formula (b), applying De Morgan's Law:\[\overline{p \wedge \bar{q}} = \bar{p} \vee q\]This is because negating an "and" statement changes it to an "or" statement with negated terms.
04

Simplify Formula (c) - Apply Distributive Law

In formula (c), distribute \(p\) across the terms in the parenthesis:\[p \wedge (\bar{p} \vee q) = (p \wedge \bar{p}) \vee (p \wedge q)\]Since \(p \wedge \bar{p}\) is false, it is eliminated, leaving:\[p \wedge q\]

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 91影视!

Key Concepts

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

Logical Implication
Logical implication is a fundamental concept in Boolean algebra. It expresses a relationship between two statements, often denoted as \( p \Rightarrow q \). This can be read as "if \( p \) then \( q \)".
Important to know is how it can be transformed logically. The implication \( p \Rightarrow q \) can be rewritten as \( \bar{p} \vee q \). This transformation uses equivalent expressions to simplify or manipulate logical formulas.
  • It can simplify reasoning about conditions.
  • The transformation changes a conditional into a disjunctive form.
By breaking down complex logical expressions into simple equivalents, logical implications help in solving and understanding different logical puzzles and equations.
De Morgan's Laws
De Morgan's Laws are rules that reveal the connection between conjunctions ("and") and disjunctions ("or") when negated. The laws are especially useful in simplifying logical expressions.
  • The first law states: \( \overline{p \wedge q} = \bar{p} \vee \bar{q} \).
  • The second law states: \( \overline{p \vee q} = \bar{p} \wedge \bar{q} \).
This means a negated "and" becomes "or" with negated terms, and a negated "or" becomes "and" with negated terms. This transformation allows for turning complex negations into simpler expressions.
In practical terms, De Morgan's Laws help when trying to simplify or refactor expressions for clearer calculation and understanding.
Distributive Law
The distributive law in Boolean algebra is quite similar to what you encounter in arithmetic. It allows the distribution of one term across terms within parentheses. The most common form used is:\[ p \wedge (q \vee r) = (p \wedge q) \vee (p \wedge r) \]
This means you can "spread" the term across the elements inside the parenthesis. The law facilitates the rearrangement and simplification of expressions.
  • Helps in expanding expressions for further simplification.
  • Useful in combining expressions efficiently.
By using the distributive law, we can systematically break down and combine logical expressions, making them easier to work with and understand.
Contradiction in Logic
Contradiction occurs when a statement is always false, no matter the values of involved variables. This is a key concept in logic as it signals an inconsistency or a fundamental error in reasoning.
A classic example is \( p \wedge \bar{p} \), meaning "\( p \) and not \( p \)", which can never be true. It鈥檚 an impossible situation.
  • Contradictions simplify formulas by identifying parts that contribute no value.
  • They help clarify logical arguments by pinpointing unsatisfiable conditions.
Recognizing contradictions is crucial in logic, as they often need elimination to streamline complex expressions to their simplest form.

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

Express each of the following compound statements symbolically: (a) The product \(x y=0\) if and only if either \(x=0\) or \(y=0\). (b) The integer \(n=4\) if and only if \(7 n-5=23\). (c) A necessary condition for \(x=2\) is \(x^{4}-x^{2}-12=0\). (d) A sufficient condition for \(x=2\) is \(x^{4}-x^{2}-12=0\). (e) For \(x^{4}-x^{2}-12=0,\) it is both sufficient and necessary to have \(x=2\). (f) The sum of squares \(x^{2}+y^{2}>1\) iff both \(x\) and \(y\) are greater than \(1 .\)

Consider the following statements: \(p: \quad\) Niagara Falls is in New York. \(q: \quad\) New York City is the state capital of New York. r: New York City will have more than 40 inches of snow in 2525 . The statement \(p\) is true, and the statement \(q\) is false. Represent each of the following statements by a formula. What is their truth value if \(r\) is true? What if \(r\) is false? (a) Niagara Falls is in New York if and only if New York City is the state capital of New York. (b) Niagara Falls is in New York iff New York City will have more than 40 inches of snow in \(2525 .\) (c) Niagara Falls is in New York or New York City is the state capital of New York if and only if New York City will have more than 40 inches of snow in \(2525 .\)

Rewrite the following expressions as conjunction: (a) \(4 \leq x \leq 7\) (b) \(4 \leq x \leq 7\) (c) \(4 \leq x<7\)

Determine whether these statements are true or false: (a) There exists an even prime integer. (b) There exist integers \(s\) and \(t\) such that \(10\). (e) For every integer \(n\), there exists an integer \(m\) such that \(m>n^{2}\).

Determine the truth values of these statements: (a) \((0 \in \mathbb{Q}) \wedge(-4 \in \mathbb{Z})\) (b) \((-4 \in \mathbb{N}) \vee(3 \in 2 \mathbb{Z})\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation
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.