Problem 48
Question
Let \(t\) be a tautology and \(p\) an arbitrary proposition. Find the truth value of each. $$(p \vee t) \rightarrow t$$
Step-by-Step Solution
Verified Answer
The truth value of the compound proposition \((p \vee t) \rightarrow t\) is always true, regardless of the truth value of the arbitrary proposition \(p\).
1Step 1: Analyze given propositions
We are given a tautology \(t\) and an arbitrary proposition \(p\).
A tautology is a proposition that is always true, regardless of the truth values of its constituent propositions. In this case, the truth value of \(t\) is true.
An arbitrary proposition \(p\) can be either true or false; we will consider both possibilities.
2Step 2: Create a truth table
To find the truth value of the compound proposition \((p \vee t) \rightarrow t\), we will create a truth table considering all possible truth values for \(p\). Since \(t\) is always true, there are only two possibilities for the truth value of \(p\) - true and false.
| \(p\) | \(t\) | \(p \vee t\) | \((p \vee t) \rightarrow t\) |
|-----|-----|------|-------|
| T | T | T | T |
| F | T | T | T |
3Step 3: Analyze the truth table
Looking at the truth table, we can see that for all possible truth values of \(p\), the compound proposition \((p \vee t) \rightarrow t\) has a truth value of true.
So, the truth value of the given compound proposition \((p \vee t) \rightarrow t\) is true, regardless of the truth value of the arbitrary proposition \(p\).
Key Concepts
Truth TableCompound PropositionPropositional Logic
Truth Table
Truth tables are a fundamental tool in propositional logic, making abstract concepts easier to understand through visual representation. A truth table lists all possible truth values of given propositions, along with the results of those values for a particular logical expression. For example, when evaluating
- \((p \vee t) \rightarrow t\), where \(t\) is always true, the truth table helps us visualize how the compound proposition behaves.
- To create a truth table for our expression, list all possible truth values of \(p\). Since \(t\) is a tautology, it's always true. Hence, there are just two possibilities: \(p\) being true (T) or false (F).
Compound Proposition
A compound proposition is formed by combining two or more simpler propositions using logical connectors such as "and" (\(\land\)), "or" (\(\lor\)), and "implies" (\(\rightarrow\)). These connectors help determine the overall truth value of the compound statement based on its components. In the example compound proposition
- \((p \vee t)\) and \((p \vee t) \rightarrow t\), the connector \(\lor\) is used in the first part, representing an 'or' operation, which results in true if at least one of \(p\) or \(t\) is true.
- Meanwhile, the implication operator \(\rightarrow\) correlates the results, implying that if the left side is true, the right side must also be true for the whole expression to be true.
Propositional Logic
Propositional logic forms the backbone of logical reasoning in mathematics and computer science. It's a branch of logic that deals with propositions, which are statements that can either be true or false. Propositions are combined to form compound propositions using logical operations. The basic logical operations include
- "and" (\(\land\)), "or" (\(\lor\)), and "not" (\(eg\)),
- "implies" (\(\rightarrow\)), which evaluates to false only when a true proposition leads to a false one.
Other exercises in this chapter
Problem 47
Let UD \(=\) set of integers, \(\mathrm{P}(x, y) : x\) is a multiple of \(y,\) and \(Q(x, y) : x \geq y\) Determine the truth value of each proposition. $$(\for
View solution Problem 47
Let UD = set of integers, \(P(x, y): x\) is a multiple of \(y,\) and \(Q(x, y): x \geq y\) Determine the truth value of each proposition. $$(\forall x)(\exists
View solution Problem 48
Let UD \(=\) set of integers, \(\mathrm{P}(x, y) : x\) is a multiple of \(y,\) and \(Q(x, y) : x \geq y\) Determine the truth value of each proposition. $$(\for
View solution Problem 48
Show that the connectives \(\wedge, \rightarrow,\) and \(\leftrightarrow\) can be expressed in terms of v and \(\sim .\) (Hint: Use Exercise 44, law 18, and Tab
View solution