Determine whether the following compound propositions are tautologies, contradictions, or neither.

1)

p↔¬p

Tautology

Contradiction

Neither a tautology nor a contradiction

2)

p→¬p

Tautology

Contradiction

Neither a tautology or a contradiction

3)

(p∧q)→p

Tautology

Contradiction.

Neither a tautology nor a contradiction.

Let's analyze each compound proposition to determine whether it is a tautology, contradiction, or neither.

### 1. \( p \leftrightarrow \neg p \)

- **Explanation**: This is a biconditional (if and only if) statement where \( p \) is true if and only if \( \neg p \) is true.
- **Truth Table**:

| \( p \) | \( \neg p \) | \( p \leftrightarrow \neg p \) |
|:------:|:-----------:|:-----------------------------:|
|   T    |      F      |               F               |
|   F    |      T      |               F               |

- **Conclusion**: The statement \( p \leftrightarrow \neg p \) is false for both possible truth values of \( p \). Hence, it is a **contradiction**.

### 2. \( p \rightarrow \neg p \)

- **Explanation**: This is an implication where \( p \) implies \( \neg p \).
- **Truth Table**:

| \( p \) | \( \neg p \) | \( p \rightarrow \neg p \) |
|:------:|:-----------:|:-------------------------:|
|   T    |      F      |             F             |
|   F    |      T      |             T             |

- **Conclusion**: The statement \( p \rightarrow \neg p \) is not true for all values of \( p \); it is only true when \( p \) is false. Therefore, it is **neither a tautology nor a contradiction**.

### 3. \( (p \wedge q) \rightarrow p \)

- **Explanation**: This is an implication where \( p \) and \( q \) together imply \( p \).
- **Truth Table**:

| \( p \) | \( q \) | \( p \wedge q \) | \( (p \wedge q) \rightarrow p \) |
|:------:|:------:|:----------------:|:--------------------------------:|
|   T    |   T    |        T         |               T                 |
|   T    |   F    |        F         |               T                 |
|   F    |   T    |        F         |               T                 |
|   F    |   F    |        F         |               T                 |

- **Conclusion**: The statement \( (p \wedge q) \rightarrow p \) is always true, regardless of the truth values of \( p \) and \( q \). Therefore, it is a **tautology**.

### Final Answers:
1. \( p \leftrightarrow \neg p \): **Contradiction**
2. \( p \rightarrow \neg p \): **Neither a tautology nor a contradiction**
3. \( (p \wedge q) \rightarrow p \): **Tautology**

Do you want further explanations on any of these, or have any questions?

Here are 5 related questions:
1. What is the difference between a tautology and a contradiction?
2. Can you explain why \( p \leftrightarrow \neg p \) is always false?
3. How does the truth table for \( p \rightarrow q \) change when both \( p \) and \( q \) are negated?
4. What would be the truth table for \( \neg (p \wedge q) \rightarrow \neg p \)?
5. How can you use De Morgan's laws to simplify logical expressions?

**Tip:** When determining whether a compound proposition is a tautology or a contradiction, constructing a truth table is a reliable method.

Understand whether compound propositions such as p ↔ ¬p, p → ¬p, and (p ∧ q) → p are tautologies, contradictions, or neither through detailed truth table analysis and logical reasoning. Explore the concepts of logic and propositional calculus with practical examples.

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