Show that `~(p harr q) -= ( p ^^ ~q) vv(~p ^^q)`.

Explanation

To show that \( \neg(p \leftrightarrow q) \equiv (p \land \neg q) \lor (\neg p \land q) \), we will use truth tables to verify the equivalence of the two expressions.

### Step 1: Define the Variables
Let’s define the variables \( p \) and \( q \). Each variable can either be true (T) or false (F). We will consider all possible combinations of truth values for \( p \) and \( q \).

### Step 2: Create a Truth Table
We will create a truth table that includes the columns for \( p \), \( q \), \( p \leftrightarrow q \), \( \neg(p \leftrightarrow q) \), \( \neg p \), \( \neg q \), \( p \land \neg q \), \( \neg p \land q \), and finally \( (p \land \neg q) \lor (\neg p \land q) \).

| \( p \) | \( q \) | \( p \leftrightarrow q \) | \( \neg(p \leftrightarrow q) \) | \( \neg p \) | \( \neg q \) | \( p \land \neg q \) | \( \neg p \land q \) | \( (p \land \neg q) \lor (\neg p \land q) \) |
|---------|---------|---------------------------|----------------------------------|--------------|--------------|----------------------|----------------------|----------------------------------------------|
| T | T | T | F | F | F | F | F | F |
| T | F | F | T | F | T | T | F | T |
| F | T | F | T | T | F | F | T | T |
| F | F | T | F | T | T | F | F | F |

### Step 3: Analyze the Truth Table
Now, we will analyze the columns for \( \neg(p \leftrightarrow q) \) and \( (p \land \neg q) \lor (\neg p \land q) \).

- The column for \( \neg(p \leftrightarrow q) \) has the values: F, T, T, F.
- The column for \( (p \land \neg q) \lor (\neg p \land q) \) has the values: F, T, T, F.

### Step 4: Conclusion
Since both columns match for all combinations of truth values, we conclude that:

\[ \neg(p \leftrightarrow q) \equiv (p \land \neg q) \lor (\neg p \land q) \]

This proves that the two expressions are logically equivalent.