The statement `(p^^(~q)) implies (p implies ~q)` is

Options

A. equivalent to` pvvq`
B. equivalent to` (~p)vv(~q)`
C. a contradiction
D. a tautology

Answer

To determine the truth of the statement \((p \land \neg q) \implies (p \implies \neg q)\), we will analyze it step by step.

### Step 1: Understand the Components
- The expression consists of two parts: \(p \land \neg q\) (the left side) and \(p \implies \neg q\) (the right side).
- The implication \(A \implies B\) is true unless \(A\) is true and \(B\) is false.

### Step 2: Analyze the Left Side
- The left side \(p \land \neg q\) is true if both \(p\) is true and \(q\) is false.
- Therefore, we will consider two cases for \(q\): true (T) and false (F).

### Step 3: Case Analysis
#### Case 1: \(q = T\)
- If \(q\) is true, then \(\neg q\) is false.
- Thus, \(p \land \neg q\) becomes \(p \land F\), which is false regardless of the value of \(p\).
- Since the left side is false, the implication \((p \land \neg q) \implies (p \implies \neg q)\) is true (a false antecedent makes the implication true).

#### Case 2: \(q = F\)
- If \(q\) is false, then \(\neg q\) is true.
- Now, \(p \land \neg q\) becomes \(p \land T\), which is true if \(p\) is true.
- If \(p\) is true, then \(p \implies \neg q\) is \(T \implies T\), which is true.
- If \(p\) is false, then \(p \implies \neg q\) is \(F \implies T\), which is also true.
- Therefore, in this case, the implication is true regardless of the value of \(p\).

### Step 4: Conclusion
- In both cases, the implication \((p \land \neg q) \implies (p \implies \neg q)\) is true.
- Hence, the statement is a tautology.

### Final Answer
The statement \((p \land \neg q) \implies (p \implies \neg q)\) is a tautology.

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