Which of the following is logically equivalent to ( p^q) ?

Options

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

Explanation

To determine which of the given options is logically equivalent to \( p \land q \), we will analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding the Original Statement**:
The original statement is \( p \land q \), which means "p and q".

2. **Option 1: \( p \implies \neg q \)**:
- We know that \( p \implies \neg q \) can be rewritten using the implication equivalence:
\[ p \implies \neg q \equiv \neg p \lor \neg q \]
- By De Morgan's Law, \( \neg p \lor \neg q \) can be expressed as:
\[ \neg (p \land q) \]
- Thus, \( p \implies \neg q \) is not equivalent to \( p \land q \).

3. **Option 2: \( \neg p \lor \neg q \)**:
- This option is already in the form of \( \neg p \lor \neg q \).
- Again, by De Morgan's Law, we have:
\[ \neg p \lor \neg q \equiv \neg (p \land q) \]
- Therefore, \( \neg p \lor \neg q \) is also not equivalent to \( p \land q \).

4. **Option 3: \( \neg p \implies \neg q \)**:
- Using the implication equivalence, we can rewrite this as:
\[ \neg p \implies \neg q \equiv p \lor \neg q \]
- However, we need to analyze this further. We can express \( \neg p \implies \neg q \) as:
\[ \neg p \lor \neg q \]
- By De Morgan's Law, we can rewrite it as:
\[ \neg (p \land q) \]
- This means \( \neg p \implies \neg q \) is not equivalent to \( p \land q \).

5. **Option 4: \( \neg \neg p \land \neg q \)**:
- Simplifying this gives us:
\[ p \land \neg q \]
- This is not equivalent to \( p \land q \).

### Conclusion:
After analyzing all the options, we find that none of the options provided are logically equivalent to \( p \land q \).

### Final Answer:
None of the options are logically equivalent to \( p \land q \).