Which of the following statement is not a fallacy?

Options

A. `p^^(~(~p rArr q))`
B. `~((p ^^ q) rArr p)`
C. `~(p rArr (p vvq))`
D. `~p vv (~p rArr q)`

Answer

To determine which of the given statements is not a fallacy, we need to analyze each option using truth tables and logical operations. A statement is considered a fallacy if it is always false, meaning that all truth values yield false. Let's go through the options step by step.

### Step 1: Analyze Option 1
**Statement:** \( P \cap \neg(\neg P) \implies Q \)

1. **Construct the truth table:**
- Let \( P \) and \( Q \) take values: True (T) and False (F).
- The truth values for \( P \) and \( Q \) are:
- \( P: T, T, F, F \)
- \( Q: T, F, T, F \)

2. **Calculate \( \neg P \):**
- \( \neg P: F, F, T, T \)

3. **Calculate \( \neg(\neg P) \):**
- \( \neg(\neg P): T, T, F, F \)

4. **Calculate \( P \cap \neg(\neg P) \):**
- \( P \cap \neg(\neg P): T, T, F, F \) (This is just \( P \))

5. **Calculate \( P \cap \neg(\neg P) \implies Q \):**
- The implication \( P \implies Q \) is true unless \( P \) is true and \( Q \) is false.
- The results are: T (T implies T), F (T implies F), F (F implies T), T (F implies F).
- So, the truth values are: T, F, T, T.

6. **Negate the entire expression:**
- Negating gives: F, T, F, F.

**Conclusion for Option 1:** Not all values are false, hence it is not a fallacy.

### Step 2: Analyze Option 2
**Statement:** \( P \cap Q \implies P \)

1. **Construct the truth table:**
- \( P: T, T, F, F \)
- \( Q: T, F, T, F \)

2. **Calculate \( P \cap Q \):**
- \( P \cap Q: T, F, F, F \)

3. **Calculate \( P \cap Q \implies P \):**
- The results are: T (T implies T), F (F implies T), T (F implies F), T (F implies F).
- So, the truth values are: T, T, T, T.

4. **Negate the entire expression:**
- Negating gives: F, F, F, F.

**Conclusion for Option 2:** All values are false, hence it is a fallacy.

### Step 3: Analyze Option 3
**Statement:** \( P \cup Q \implies P \cap Q \)

1. **Construct the truth table:**
- \( P: T, T, F, F \)
- \( Q: T, F, T, F \)

2. **Calculate \( P \cup Q \):**
- \( P \cup Q: T, T, T, F \)

3. **Calculate \( P \cap Q \):**
- \( P \cap Q: T, F, F, F \)

4. **Calculate \( P \cup Q \implies P \cap Q \):**
- The results are: T (T implies T), F (T implies F), T (T implies F), T (F implies F).
- So, the truth values are: T, F, T, T.

5. **Negate the entire expression:**
- Negating gives: F, T, F, F.

**Conclusion for Option 3:** Not all values are false, hence it is not a fallacy.

### Step 4: Analyze Option 4
**Statement:** \( \neg P \implies Q \)

1. **Construct the truth table:**
- \( P: T, T, F, F \)
- \( Q: T, F, T, F \)

2. **Calculate \( \neg P \):**
- \( \neg P: F, F, T, T \)

3. **Calculate \( \neg P \implies Q \):**
- The results are: T (F implies T), F (F implies F), T (T implies T), F (T implies F).
- So, the truth values are: T, F, T, F.

4. **Negate the entire expression:**
- Negating gives: F, T, F, T.

**Conclusion for Option 4:** Not all values are false, hence it is not a fallacy.

### Final Conclusion
After analyzing all four options, we find that Option 2 is the only statement that is a fallacy (all values are false). Therefore, the answer to the question "Which of the following statement is not a fallacy?" is Options 1, 3, and 4.