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Let's construct the truth tables for the given compound statements step by step.  

### Given statements:
1. **(a)** \( \neg [\neg p \land \neg q] \)
2. **(b)** \( (p \land q) \land \neg r \)
3. **(c)** \( \neg [\neg p \lor \neg q] \)

---

### **Step 1: Identify the variables**
The statements involve three logical variables:  
- \( p \)  
- \( q \)  
- \( r \) (only in part b)

Each variable can be **True (T)** or **False (F)**.

---

### **Truth Table for (a): \( \neg [\neg p \land \neg q] \)**

| \( p \) | \( q \) | \( \neg p \) | \( \neg q \) | \( \neg p \land \neg q \) | \( \neg [\neg p \land \neg q] \) |
|:-:|:-:|:-:|:-:|:-:|:-:|
| T | T | F | F | F | T |
| T | F | F | T | F | T |
| F | T | T | F | F | T |
| F | F | T | T | T | F |

Thus, the final result for (a) is equivalent to **\( p \lor q \) (Disjunction of \( p \) and \( q \))**.

---

### **Truth Table for (b): \( (p \land q) \land \neg r \)**

| \( p \) | \( q \) | \( r \) | \( p \land q \) | \( \neg r \) | \( (p \land q) \land \neg r \) |
|:-:|:-:|:-:|:-:|:-:|:-:|
| T | T | T | T | F | F |
| T | T | F | T | T | T |
| T | F | T | F | F | F |
| T | F | F | F | T | F |
| F | T | T | F | F | F |
| F | T | F | F | T | F |
| F | F | T | F | F | F |
| F | F | F | F | T | F |

Thus, the final result for (b) is **True** only when \( p \) and \( q \) are both True, and \( r \) is False.

---

### **Truth Table for (c): \( \neg [\neg p \lor \neg q] \)**

| \( p \) | \( q \) | \( \neg p \) | \( \neg q \) | \( \neg p \lor \neg q \) | \( \neg [\neg p \lor \neg q] \) |
|:-:|:-:|:-:|:-:|:-:|:-:|
| T | T | F | F | F | T |
| T | F | F | T | T | F |
| F | T | T | F | T | F |
| F | F | T | T | T | F |

Thus, the final result for (c) is equivalent to **\( p \land q \) (Conjunction of \( p \) and \( q \))**.

---

### **Final Conclusions**
1. **(a):** \( \neg [\neg p \land \neg q] \equiv p \lor q \)  
2. **(b):** \( (p \land q) \land \neg r \)  
3. **(c):** \( \neg [\neg p \lor \neg q] \equiv p \land q \)  

Do you want a further breakdown or any clarifications?  

### **Related Questions**
1. How does De Morgan’s theorem help simplify (a) and (c)?  
2. What are the real-life applications of logical conjunction and disjunction?  
3. How does the truth table method help in Boolean algebra simplifications?  
4. Can you derive the logical equivalence of these statements without using truth tables?  
5. How does statement (b) relate to conditional logic?  

### **Tip:**  
Truth tables are a powerful tool for verifying logical expressions. If a statement is always true, it is a **tautology**; if always false, it is a **contradiction**.

Construct the truth table for the compound statement: (a) ~[~p∧~q] (b) (p∧q)∧~r (c) ~[~p∨~q]

Learn how to construct truth tables for complex logical expressions like ~[~p∧~q], (p∧q)∧~r, and ~[~p∨~q]. This guide covers essential concepts of logic including De Morgan's Law, conjunction, disjunction, and negation. Suitable for high school students (Grades 9-12).

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