The compound statement `(prarr ~q)vv(p^^q)` is logically equivalent to

Options

A. `phArr q`
B. `p^^q`
C. Tautology
D. Contradication

Answer

To solve the compound statement \( (p \rightarrow \neg q) \lor (p \land q) \) and determine its logical equivalence, we will construct a truth table.

### Step 1: Define the truth values for \( p \) and \( q \)

We will consider all possible truth values for \( p \) and \( q \):
- \( p \): True (T), True (T), False (F), False (F)
- \( q \): True (T), False (F), True (T), False (F)

### Step 2: Calculate \( \neg q \)

The negation of \( q \) will be:
- If \( q \) is T, then \( \neg q \) is F.
- If \( q \) is F, then \( \neg q \) is T.

So, we have:
- \( \neg q \): F, T, F, T

### Step 3: Calculate \( p \rightarrow \neg q \)

The implication \( p \rightarrow \neg q \) is defined as:
- True unless \( p \) is True and \( \neg q \) is False.

Calculating this:
- For \( (T \rightarrow F) \): False
- For \( (T \rightarrow T) \): True
- For \( (F \rightarrow F) \): True
- For \( (F \rightarrow T) \): True

So, we have:
- \( p \rightarrow \neg q \): F, T, T, T

### Step 4: Calculate \( p \land q \)

The conjunction \( p \land q \) is True only if both \( p \) and \( q \) are True:
- For \( (T \land T) \): True
- For \( (T \land F) \): False
- For \( (F \land T) \): False
- For \( (F \land F) \): False

So, we have:
- \( p \land q \): T, F, F, F

### Step 5: Calculate \( (p \rightarrow \neg q) \lor (p \land q) \)

Now we combine the results from steps 3 and 4 using the disjunction \( \lor \):
- \( (F \lor T) \): True
- \( (T \lor F) \): True
- \( (T \lor F) \): True
- \( (T \lor F) \): True

So, we have:
- \( (p \rightarrow \neg q) \lor (p \land q) \): T, T, T, T

### Conclusion

Since the final column of the truth table is all True, the compound statement \( (p \rightarrow \neg q) \lor (p \land q) \) is logically equivalent to a tautology.

### Final Answer

The compound statement \( (p \rightarrow \neg q) \lor (p \land q) \) is logically equivalent to **Tautology**.