Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Construct the truth table for $$ (p \wedge(p \rightarrow q) \wedge(q \rightarrow r)) \rightarrow r $$ Simplify this expression to one using only \(\wedge, \vee,\) and \(\neg\)

Short Answer

Expert verified
The truth table has eight rows; the expression simplifies to \((p \wedge (\neg p \vee q) \wedge (\neg q \vee r)) \vee r\).

Step by step solution

01

List All Variables and Possible Combinations

The variables present in the expression are \(p, q,\) and \(r\). Since each can be true (T) or false (F), we have \(2^3 = 8\) possible combinations. We will list these combinations as rows in the truth table.
02

Compute Sub-expressions

Start by computing the inner expressions separately for each row of possibilities:- \( (p \rightarrow q) \) using the logical implication: it's false only when \( p \) is true and \( q \) is false.- \( (q \rightarrow r) \) similarly, it's false only when \( q \) is true and \( r \) is false.
03

Evaluate Compound Sub-expressions

Compute the compound expressions:- \( p \wedge (p \rightarrow q) \) using logical AND: it's true only when both \( p \) and \( (p \rightarrow q) \) are true.- Next, compute \((p \wedge (p \rightarrow q) \wedge (q \rightarrow r))\) which is true only when all three components are true.
04

Evaluate the Main Expression

Finally, evaluate the main expression, \((p \wedge(p \rightarrow q) \wedge(q \rightarrow r)) \rightarrow r\), using the logical implication. This expression is false only when the left part is true and \( r \) is false.
05

Simplify Using Logical Equivalences

To simplify, notice that if all conditions in the antecedent \((p \wedge (p \rightarrow q) \wedge (q \rightarrow r))\) lead to the conclusion \(r\), then the entire expression can be represented through "or" and "not" operators:- By properties of implications: \( p \rightarrow q \equiv eg p \vee q \) and \( q \rightarrow r \equiv eg q \vee r \).- Rewriting the statement, we have \((p \wedge (eg p \vee q) \wedge (eg q \vee r)) \rightarrow r \).- The final simplification of \( (p \wedge (eg p \vee q) \wedge(eg q \vee r)) \vee r \).

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 Vaia!

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 logic known as "if...then" statements. In mathematics and logic, it is denoted by the symbol \( \rightarrow \). A statement of the form \( p \rightarrow q \) means "if \( p \) then \( q \)." It implies that if \( p \) is true, \( q \) must also be true for the entire expression to hold. The logical implication is only false when \( p \) is true and \( q \) is false.
In the context of constructing a truth table, you need to evaluate every possible situation for the variables involved. For example, in \( (p \rightarrow q) \), the truth depends on the combination of truth values of \( p \) and \( q \). Remember, the only case where \( p \rightarrow q \) is false is when \( p \) is true and \( q \) is false.
When simplifying expressions, implications can be rewritten using "or" and "not". For instance, \( p \rightarrow q \) is logically equivalent to \( eg p \vee q \). This transformation can often help simplify complex logical statements.
Logical AND
The logical AND is represented by the symbol \( \wedge \) and is used to combine statements that are both required to be true. It is similar to the word "and" in regular language, but with a strict requirement that both operands must be true for the result to be true. In a truth table, the \( p \wedge q \) expression is only true when both \( p \) and \( q \) are true.
For example, if you have expressions like \( p \wedge (p \rightarrow q) \), you are looking at a situation where both conditions need to be satisfied. You calculate the truth value of each individual component first, and then combine them using AND logic.
This characteristic of AND makes it particularly useful in checking combinations where multiple conditions are minimum requirements. In logical expressions, if any component joined by AND is false, the whole expression results false, ensuring a powerful filtering criteria for logical operations.
Logical OR
Logical OR is denoted by the symbol \( \vee \). It combines two statements where only one of them needs to be true for the entire expression to be true. This logical operation resembles the casual "or" in English, but it also includes the situation where both conditions are true as satisfactory.
In a truth table, \( p \vee q \) becomes true if either \( p \) or \( q \), or both, are true. It is only false when both \( p \) and \( q \) are false.
Using Logical OR can simplify expressions, especially in combination with logical NOT. For example, an implication \( p \rightarrow q \) can be rewritten as \( eg p \vee q \), showing the power of OR in logical operations. This transformation is crucial for simplifying logical expressions into more manageable forms using basic operations of AND, OR, and NOT.
Logical Equivalence
Logical equivalence occurs when two statements are always equal or have the same truth value in all possible scenarios. They are represented by the symbol \( \equiv \). For logical expressions, equivalence is essential in simplifying complex logical statements.
Consider the simplification of a truth table expression. Using logical equivalence, you can transform complex expressions into simpler forms by using known equivalences. For instance, \( p \rightarrow q \equiv eg p \vee q \) is a basic logic equivalence that helps reduce implications to simpler forms.
Another example involves using equivalences to combine operations: an expression like \( (p \wedge (eg p \vee q)) \) can be simplified using distributive laws into something more simplified. Logical equivalence thus helps to not only evaluate but also rewrite logical expressions into basic components of AND, OR, and NOT, maintaining the truth value across the transformation.

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

Find at least two different ways to fill in the ellipses in the set descriptions given. For example, \(\\{2,4, \ldots, 12\\}\) could be written either \(\mid 2 n: 1 \leq n \leq 6\) and \(n \in \mathbb{N})\) or \(\mid n+1: n \in\\{1,3,5,7,11\\}\\}\) (a) \([1,3, \ldots, 31)\) (b) \(\\{1,2, \ldots, 26 \mid\) (c) \(\\{2,5, \ldots, 32\\}\)

(a) Find the resolvant of \((p \vee q)\) and \((\neg p \vee r)\) on \(p\). (b) Find the resolvant of \((p \vee q \vee r \vee s)\) and \((\neg p \vee \neg q \vee t)\) on \(p\). (c) Find the resolvant of \((p \vee q)\) and \(\neg p\) on \(p\). (d) Find the resolvant of \((p)\) and \((\neg p)\) on \(p\). (e) Which resolvant above from parts (a) through (d) is a tautology? Which is tautologically false?

This problem concerns the following six sets: $$\begin{array}{c}A=\\{0,2,4,6\\} \quad B=(1,3,5) \quad C=\\{0,1,2,3,4,5,6,7\\} \\\D=\emptyset \quad E=\mathbb{N} \quad F=\\{10,2,4,6\\} \mid\end{array}$$ (a) What sets are subsets of \(A\) ? (b) What sets are subsets of \(B\) ? (c) What sets are subsets of \(C\) ? (d) What sets are subsets of \(D\) ? (c) What sets are subsets of \(E\) ? (f) What sets are subsets of \(F\) ?

Find a CNF for each of the following formulas, and prove that each formula is a tautology. (a) \((p \wedge p) \leftrightarrow p\) (b) \((p \wedge(p \rightarrow q)) \rightarrow q\) (c) \((p \rightarrow(r \rightarrow q)) \leftrightarrow((p \wedge r) \rightarrow q)\) (d) \((p \rightarrow r) \leftrightarrow(\neg r \rightarrow \neg p)\)

Let \(\phi=(p \rightarrow q) \rightarrow((r \wedge \neg s) \rightarrow q)\). For each of the following interpretations of \(p, q, r,\) and \(s,\) compute \(I(\phi)\) using the truth tables for \(\neg, v, \wedge, \rightarrow,\) and \(\leftrightarrow\) (a) \(I(p)=T, I(q)=T, I(r)=F,\) and \(I(s)=T\) (b) \(I(p)=T, I(q)=F, I(r)=T,\) and \(I(s)=F\) (c) \(I(p)=F, I(q)=T, I(r)=T,\) and \(I(s)=F\) (d) \(I(p)=F, I(q)=F, I(r)=T,\) and \(I(s)=F\)

See all solutions

Recommended explanations on Computer Science Textbooks

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.

Sign-up for free