https://file.aiquickdraw.com/mathbot/file/1726969517164vqqy66kg.jpg

The problem asks to prove that the given logical statement is a tautology, meaning it is always true regardless of the truth values of its variables. The formula from the image is:

\[
(((P \vee Q) \rightarrow R) \vee (\neg P)) \rightarrow (Q \rightarrow R)
\]

To prove this is a tautology, we can follow a truth table approach or logical equivalence. Let's break this down:

### Step 1: Simplify the Expression
We will break the formula into smaller components and determine if the logical equivalences hold:

- \( P \vee Q \): This is the logical "or" between \( P \) and \( Q \).
- \( (P \vee Q) \rightarrow R \): The implication that if \( P \) or \( Q \) is true, then \( R \) must be true.
- \( \neg P \): The negation of \( P \).
- \( ((P \vee Q) \rightarrow R) \vee (\neg P) \): This states that either the implication from \( (P \vee Q) \rightarrow R \) is true or \( \neg P \) is true.
- \( Q \rightarrow R \): This is a simple implication between \( Q \) and \( R \).
- Finally, the overall expression is an implication: \( (((P \vee Q) \rightarrow R) \vee (\neg P)) \rightarrow (Q \rightarrow R) \).

### Step 2: Create the Truth Table
We will now construct a truth table with all possible truth values for \( P \), \( Q \), and \( R \), and evaluate the expression step by step.

Let me calculate this for you.Here is the truth table for the expression:

\[
(((P \vee Q) \rightarrow R) \vee (\neg P)) \rightarrow (Q \rightarrow R)
\]

| \( P \)  | \( Q \)  | \( R \)  | \( (((P \vee Q) \rightarrow R) \vee \neg P) \rightarrow (Q \rightarrow R) \) |
|:-------:|:-------:|:-------:|:-------------------------------------------------:|
| False   | False   | False   | True  |
| False   | False   | True    | True  |
| False   | True    | False   | False |
| False   | True    | True    | True  |
| True    | False   | False   | True  |
| True    | False   | True    | True  |
| True    | True    | False   | True  |
| True    | True    | True    | True  |

From the table, we can see that the expression evaluates to **True** in all but one case. Since one case produces **False**, the statement is **not a tautology**.

Do you want any clarifications on this result?

Here are 5 follow-up questions to consider:
1. What is a counterexample for this expression?
2. How can truth tables be used to analyze other logical expressions?
3. Can De Morgan's laws help simplify parts of this logical expression?
4. How would changing the expression to \( P \rightarrow R \) affect the result?
5. How can you interpret implications in formal logic in practical reasoning?

**Tip:** When working with logical implications, remember that \( P \rightarrow Q \) is equivalent to \( \neg P \vee Q \).

Prove that the following statement formula is a tautology: (((P ∨ Q) → R) ∨ ¬P) → (Q → R)

Learn how to prove that the logical formula (((P ∨ Q) → R) ∨ ¬P) → (Q → R) is a tautology using truth tables and logical equivalences. This problem involves key concepts in propositional logic such as implication, disjunction, and negation, and is suitable for college-level students.

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