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Construct a truth table for \(((p \rightarrow q) \rightarrow r) \rightarrow s\)

Short Answer

Expert verified
Create columns for all variables and intermediary steps. Fill in truth values for all combinations and compute step by step.

Step by step solution

01

- List All Possible Combinations of Truth Values

For variables \(p\), \(q\), \(r\), and \(s\), list all possible combinations of their truth values (True and False).
02

- Calculate \(p \rightarrow q\)

Use the implication rule: \(p \rightarrow q\) is False only if \(p\) is True and \(q\) is False; otherwise, it is True. Compute \(p \rightarrow q\) for each combination from Step 1.
03

- Calculate \((p \rightarrow q) \rightarrow r\)

Use the results from Step 2 to determine \((p \rightarrow q) \rightarrow r\). Apply the implication rule again.
04

- Calculate \(((p \rightarrow q) \rightarrow r) \rightarrow s\)

Take the results from Step 3 and use them to compute \(((p \rightarrow q) \rightarrow r) \rightarrow s\). Apply the implication rule one more time.
05

- Compile the Truth Table

Create a table that includes columns for \(p\), \(q\), \(r\), \(s\), \(p \rightarrow q\), \((p \rightarrow q) \rightarrow r\), and \(((p \rightarrow q) \rightarrow r) \rightarrow s\), and fill in the calculated values from Steps 2, 3, and 4.

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Key Concepts

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

Logical Implications
In logic, an *implication* is a statement formed by two propositions. It takes the form *if p, then q* and is written as \( p \rightarrow q \).
Let's dive into how this works:
  • *If p is True* and *q is True* - the statement \(p \rightarrow q\) is True.
  • *If p is True* and *q is False* - the statement \(p \rightarrow q\) is False.
  • *If p is False* (regardless of q's value) - the statement \(p \rightarrow q\) is always True.
This might look odd, but it's because implication focuses on whether we can find any counterexamples.
If none exist, then the implication is considered true. We'll keep using this rule in each step of building the truth table.
Truth Values
Truth values are simply whether a statement is True (T) or False (F).
In logical expressions, these values help us understand the result of combining different propositions.
For our expression \(((p \rightarrow q) \rightarrow r) \rightarrow s\), we have four variables - \( p, q, r, s\).
Each can be either True or False.
Here's how we approach it:
  • We start by listing all possible combinations of these values. There are 16 combinations for 4 variables.
  • Then, we compute the truth values step by step using the rules for implications.
This systematic method ensures that we cover every possible scenario, which is the essence of creating a truth table.
Step-by-Step Process
Building a truth table is like solving a puzzle. Let's see exactly how the steps come together:
Step 1: List all possible combinations of truth values for \( p, q, r, \) and \ s\.

Step 2: Calculate \( p \rightarrow q \) using the implication rule.
This is done for each combination.

Step 3: Use the results from Step 2 to find \( (p \rightarrow q) \rightarrow r \).
Apply the implication rule again here.

Step 4: Now calculate \(((p \rightarrow q) \rightarrow r) \rightarrow s\).
We take the values from Step 3 and apply the rule one more time.

Step 5: Finally, put all these values into a table.
The columns should include all your propositions and the intermediate results.
This visual layout makes it clear how the final truth values are computed based on each step.
Utilize bullet points and a neat structure to break down each step for easy understanding.

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Most popular questions from this chapter

Find a common domain for the variables \(x, y, z,\) and \(w\) for which the statement \(\forall x \forall y \forall z \exists w((w \neq x) \wedge\) \((w \neq y) \wedge(w \neq z) )\) is true and another common domain for these variables for which it is false.

Suppose that \(a\) and \(b\) are odd integers with \(a \neq b .\) Show there is a unique integer \(c\) such that \(|a-c|=|b-c|\)

Let \(Q(x)\) be the statement " \(x+1>2 x\) . If the domain consists of all integers, what are these truth values? $$ \begin{array}{llll}{\text { a) }} & {Q(0)} & {\text { b) } Q(-1)} & {\text { c) }} \quad {Q(1)} \\ {\text { d) }} & {\exists x Q(x)} & {\text { e) } \quad \forall x Q(x)} & {\text { f) } \quad \exists x \neg Q(x)}\end{array} $$ g) \(\quad \forall x \neg Q(x)\)

For each of these arguments determine whether the argument is correct or incorrect and explain why. a) All students in this class understand logic. Xavier is a student in this class. Therefore, Xavier understands logic. b) Every computer science major takes discrete math- ematics. Natasha is taking discrete mathematics. Therefore, Natasha is a computer science major. c) All parrots like fruit. My pet bird is not a parrot. Therefore, my pet bird does not like fruit. d) Everyone who eats granola every day is healthy. Linda is not healthy. Therefore, Linda does not eagranola every day.

Exercises \(61-64\) are based on questions found in the book Symbolic Logic by Lewis Carroll. Let P(x), Q(x), R(x), and S(x) be the statements 鈥渪 is a baby,鈥 鈥渪 is logical,鈥 鈥渪 is able to manage a crocodile,鈥 and 鈥渪 is despised,鈥 respectively. Suppose that the domain consists of all people. Express each of these statements using quantifiers; logical connectives; and P(x), Q(x), R(x), and S(x). a) Babies are illogical. b) Nobody is despised who can manage a crocodile. c) Illogical persons are despised. d) Babies cannot manage crocodiles. e) Does (d) follow from (a), (b), and (c)? If not, is there a correct conclusion?

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