The logically equivalent proposition of `p harr q` is

Options

A. `(p wedge q) vv (p wedge q)`
B. `(p rarr q) wedge (q rarr p)`
C. `(p wedge q) vee (q rarr p)`
D. `(p wedge q) rarr (q vee p)`

Answer

To find the logically equivalent proposition of \( p \text{ biconditional } q \) (denoted as \( p \harr q \)), we can follow these steps:

### Step 1: Understand the biconditional statement
The biconditional statement \( p \harr q \) means that \( p \) is true if and only if \( q \) is true. This can be expressed in terms of implications:
\[ p \harr q \equiv (p \implies q) \land (q \implies p) \]

### Step 2: Create a truth table
To verify this equivalence, we can create a truth table for \( p \), \( q \), \( p \implies q \), \( q \implies p \), and \( p \harr q \).

| \( p \) | \( q \) | \( p \implies q \) | \( q \implies p \) | \( p \harr q \) |
|---------|---------|---------------------|---------------------|------------------|
| T | T | T | T | T |
| T | F | F | T | F |
| F | T | T | F | F |
| F | F | T | T | T |

### Step 3: Analyze the truth table
From the truth table, we can see:
- \( p \harr q \) is true when both \( p \) and \( q \) are true, or both are false.
- \( p \implies q \) is false only when \( p \) is true and \( q \) is false.
- \( q \implies p \) is false only when \( q \) is true and \( p \) is false.

Thus, the conjunction \( (p \implies q) \land (q \implies p) \) is true in the same cases as \( p \harr q \).

### Step 4: Conclusion
Therefore, we conclude that:
\[ p \harr q \equiv (p \implies q) \land (q \implies p) \]

### Final Answer
The logically equivalent proposition of \( p \harr q \) is \( (p \implies q) \land (q \implies p) \).

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