---### Determining if a Statement is a Tautology, Contradiction, or Neither**Instructions:**1. Complete the following truth table. Use T for true and F for false. You may add more columns, but those added columns will not be graded.**Given Statement:**~(p ∨ q) ∧ (p ∧ ~q)#### Truth Table:| p | q | ~(p ∨ q) | p ∧ ~q | ~(p ∨ q) ∧ (p ∧ ~q) ||---|---|---------|-----|------------------------|| T | T | F | F | F || T | F | F | T | F || F | T | F | F | F || F | F | T | F | F |(b) Is the statement ~(p ∨ q) ∧ (p ∧ ~q) a tautology, a contradiction, or neither? Why? Choose the best answer.- The statement is a tautology. This is because it is true for all possible true-false combinations of p and q.- The statement is a contradiction. This is because it is false for every possible combination of p and q.- The statement is neither a tautology nor a contradiction. This is since it is true for some combinations of p and q and false for others.**Answer:**- The statement is a contradiction. This is because it is false for every possible combination of p and q.**Explanation:**The truth table shows that no matter what values of p and q are chosen (both true, both false, one true and one false), the resulting expression ~(p ∨ q) ∧ (p ∧ ~q) always evaluates to false, hence it is a contradiction.[Check Button]© 2023 McGraw Hill LLC. All Rights Reserved.--- ### Determining if a Statement is a Tautology, Contradiction, or NeitherIn logic, it is important to determine whether a given statement is a tautology, a contradiction, or neither. This understanding helps in evaluating logical arguments and constructing proofs. #### Example Problem:**Given Statement:**Determine if the following statement is a tautology, a contradiction, or neither, and explain why: \[(p \vee q) \land (p \land \neg q)\]#### Options:1. **The statement is a tautology. This is because it is true for all possible true-false combinations of $p$ and $q$.**2. **The statement is a tautology. This is because it is true for some true-false combinations of $p$ and $q$ and false for others.**3. **The statement is a contradiction. This is because it is false for all possible true-false combinations of $p$ and $q$.**4. **The statement is a contradiction. This is because it is true for some true-false combinations of $p$ and $q$ and false for others.**5. **The statement is neither a tautology nor a contradiction.**#### Explanation:- **Tautology:** A statement that is always true regardless of the truth values of the propositions involved.- **Contradiction:** A statement that is always false regardless of the truth values of the propositions involved.- **Neither:** A statement that is true for some truth values and false for others.#### Graphs and Diagrams:1. **Truth Table Construction:** - **Step 1:** Evaluate $p \vee q$ - **Step 2:** Evaluate $p \land \neg q$ - **Step 3:** Combine the results using $(p \vee q) \land (p \land \neg q)$| $p$ | $q$ | $\neg q$ | $p \vee q$ | $p \land \neg q$ | $(p \vee q) \land (p \land \neg q)$ ||------|------|-------------|-------------|-------------------|------------------------------------|| T | T | F | T | F | F || T | F | T

We know thatThe following logic tablepq~pTTFTTTFFFTFTTFTFFTFFHereWe have to complete the given truth…

(a) Complete the following truth table. Use T for true and F for false. You may add more columns, but those added columns will not be graded. P 9 TT T F FT LL F F ~(pvq)^(p^~qg) 10 0 0 10 Р ~O ローロ X 9 A ローロ OVO S hs

(a) Complete the following truth table. Use T for true and F for false. You may add more columns, but those added columns will not be graded. P 9 TT T F FT LL F F ~(pvq)^(p^~qg) 10 0 0 10 Р ~O ローロ X 9 A ローロ OVO S hs

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

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### Determining if a Statement is a Tautology, Contradiction, or Neither

**Instructions:**
1. Complete the following truth table. Use T for true and F for false. You may add more columns, but those added columns will not be graded.

**Given Statement:**
~(p ∨ q) ∧ (p ∧ ~q)

#### Truth Table:

| p | q | ~(p ∨ q) | p ∧ ~q | ~(p ∨ q) ∧ (p ∧ ~q) |
|---|---|---------|-----|------------------------|
| T | T | F | F | F |
| T | F | F | T | F |
| F | T | F | F | F |
| F | F | T | F | F |

(b) Is the statement ~(p ∨ q) ∧ (p ∧ ~q) a tautology, a contradiction, or neither? Why? Choose the best answer.

- The statement is a tautology. This is because it is true for all possible true-false combinations of p and q.
- The statement is a contradiction. This is because it is false for every possible combination of p and q.
- The statement is neither a tautology nor a contradiction. This is since it is true for some combinations of p and q and false for others.

**Answer:**
- The statement is a contradiction. This is because it is false for every possible combination of p and q.

**Explanation:**
The truth table shows that no matter what values of p and q are chosen (both true, both false, one true and one false), the resulting expression ~(p ∨ q) ∧ (p ∧ ~q) always evaluates to false, hence it is a contradiction.

[Check Button]

© 2023 McGraw Hill LLC. All Rights Reserved.

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### Determining if a Statement is a Tautology, Contradiction, or Neither

In logic, it is important to determine whether a given statement is a tautology, a contradiction, or neither. This understanding helps in evaluating logical arguments and constructing proofs.

#### Example Problem:

**Given Statement:**

Determine if the following statement is a tautology, a contradiction, or neither, and explain why:
\[(p \vee q) \land (p \land \neg q)\]

#### Options:

1. **The statement is a tautology. This is because it is true for all possible true-false combinations of $p$ and $q$.**
2. **The statement is a tautology. This is because it is true for some true-false combinations of $p$ and $q$ and false for others.**
3. **The statement is a contradiction. This is because it is false for all possible true-false combinations of $p$ and $q$.**
4. **The statement is a contradiction. This is because it is true for some true-false combinations of $p$ and $q$ and false for others.**
5. **The statement is neither a tautology nor a contradiction.**

#### Explanation:

- **Tautology:** A statement that is always true regardless of the truth values of the propositions involved.
- **Contradiction:** A statement that is always false regardless of the truth values of the propositions involved.
- **Neither:** A statement that is true for some truth values and false for others.

#### Graphs and Diagrams:

1. **Truth Table Construction:**
- **Step 1:** Evaluate $p \vee q$
- **Step 2:** Evaluate $p \land \neg q$
- **Step 3:** Combine the results using $(p \vee q) \land (p \land \neg q)$

| $p$ | $q$ | $\neg q$ | $p \vee q$ | $p \land \neg q$ | $(p \vee q) \land (p \land \neg q)$ |
|------|------|-------------|-------------|-------------------|------------------------------------|
| T | T | F | T | F | F |
| T | F | T

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