` p to q` is logically equivalent to

Options

A. `p ^^ ~ q`
B. ` ~ p to ~ q`
C. ` ( p vv ~ q)`
D. none of these

Answer

To determine what `p implies q` (denoted as \( p \rightarrow q \)) is logically equivalent to, we will analyze the truth values of the expression and compare it with other logical expressions.

### Step-by-Step Solution:

1. **Understanding Implication**:
The implication \( p \rightarrow q \) can be defined in terms of truth values:
- It is **false** only when \( p \) is **true** and \( q \) is **false**.
- In all other cases, it is **true**.

2. **Truth Table for \( p \rightarrow q \)**:
We can create a truth table for \( p \) and \( q \) to analyze the implication:
\[ \begin{array}{|c|c|c|} \hline p & q & p \rightarrow q \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \\ \hline \end{array} \]

3. **Negation of \( q \)**:
Next, we find the negation of \( q \) (denoted as \( \neg q \)):
\[ \begin{array}{|c|c|c|} \hline p & q & \neg q \\ \hline T & T & F \\ T & F & T \\ F & T & F \\ F & F & T \\ \hline \end{array} \]

4. **Checking Logical Equivalence**:
We will check the logical equivalence of \( p \rightarrow q \) with other expressions:

a. **\( p \land \neg q \)** (Conjunction):
\[ \begin{array}{|c|c|c|c|} \hline p & \neg q & p \land \neg q \\ \hline T & F & F \\ T & T & T \\ F & F & F \\ F & T & F \\ \hline \end{array} \]
This does not match \( p \rightarrow q \).

b. **\( \neg p \rightarrow \neg q \)** (Contrapositive):
\[ \begin{array}{|c|c|c|c|} \hline p & q & \neg p & \neg q & \neg p \rightarrow \neg q \\ \hline T & T & F & F & T \\ T & F & F & T & T \\ F & T & T & F & F \\ F & F & T & T & T \\ \hline \end{array} \]
This does not match \( p \rightarrow q \).

c. **\( p \lor \neg q \)** (Disjunction):
\[ \begin{array}{|c|c|c|c|} \hline p & \neg q & p \lor \neg q \\ \hline T & F & T \\ T & T & T \\ F & F & F \\ F & T & T \\ \hline \end{array} \]
This does not match \( p \rightarrow q \).

5. **Final Conclusion**:
After checking all combinations, we find that \( p \rightarrow q \) is logically equivalent to \( \neg p \lor q \) (not \( p \) or \( q \)). This can be confirmed by constructing the truth table for \( \neg p \lor q \) and comparing it with \( p \rightarrow q \).

### Final Answer:
Thus, \( p \rightarrow q \) is logically equivalent to \( \neg p \lor q \).