Which of the following statement is not a fallacy? Option1 p ∧ ( ~ ( ~ p ⇒ q ) ) Option2 ~ ( ( p ∧ q ) ⇒ p ) Option3 ~ ( p ⇒ ( p ∨ q ) ) Option4 ~ p ∨ ( ~ p ⇒ q )

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 will analyze each option step by step.

### Step 1: Understanding Fallacies
A fallacy is a compound statement that is always false. We will denote a fallacy by \( F \).

### Step 2: Analyzing Each Option

#### Option 1: \( p \land \sim(\sim p \Rightarrow q) \)

1. The expression \( \sim p \Rightarrow q \) can be rewritten using the implication equivalence:
\[ \sim p \Rightarrow q \equiv p \lor q \]
Thus, \( \sim(\sim p \Rightarrow q) \equiv \sim(p \lor q) \equiv \sim p \land \sim q \).

2. Now substituting back into the original expression:
\[ p \land \sim(\sim p \Rightarrow q) \equiv p \land (\sim p \land \sim q) \equiv (p \land \sim p) \land \sim q \]
Since \( p \land \sim p \) is always false, this entire expression is a fallacy.

#### Option 2: \( \sim((p \land q) \Rightarrow p) \)

1. The expression \( (p \land q) \Rightarrow p \) can be rewritten as:
\[ \sim(p \land q) \lor p \]
Thus, \( \sim((p \land q) \Rightarrow p) \equiv \sim(\sim(p \land q) \lor p) \equiv (p \land q) \land \sim p \).

2. This simplifies to:
\[ (p \land q) \land \sim p \equiv (p \land \sim p) \land q \]
Again, since \( p \land \sim p \) is always false, this expression is also a fallacy.

#### Option 3: \( \sim(p \Rightarrow (p \lor q)) \)

1. The expression \( p \Rightarrow (p \lor q) \) is always true because if \( p \) is true, \( p \lor q \) is also true.

2. Therefore, \( \sim(p \Rightarrow (p \lor q)) \equiv \sim \text{True} \equiv \text{False} \), which is a fallacy.

#### Option 4: \( \sim p \lor (\sim p \Rightarrow q) \)

1. The expression \( \sim p \Rightarrow q \) can be rewritten as:
\[ \sim p \Rightarrow q \equiv p \lor q \]
Thus, the expression becomes:
\[ \sim p \lor (p \lor q) \equiv (\sim p \lor p) \lor q \]
Since \( \sim p \lor p \) is always true (tautology), this entire expression is also true regardless of the value of \( q \).

### Conclusion
From the analysis:
- Options 1, 2, and 3 are fallacies.
- Option 4 is not a fallacy (it is a tautology).

Thus, the answer is **Option 4**.