Using truth table show that `- (p vv q) vv (~ p ^^ q ) ` is logically equivalent to ~ p.

Answer

To show that \(- (p \lor q) \lor (\neg p \land q)\) is logically equivalent to \(\neg p\) using a truth table, we will follow these steps:

### Step 1: Create the Truth Table
We will create a truth table that includes all possible truth values for \(p\) and \(q\). There are four combinations of truth values for two variables.

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

### Step 2: Fill in the Columns
1. **Columns for \(p\) and \(q\)**: We list all combinations of truth values for \(p\) and \(q\).
2. **Column for \(p \lor q\)**: This column is true if either \(p\) or \(q\) is true.
3. **Column for \(\neg(p \lor q)\)**: This column is the negation of the previous column.
4. **Column for \(\neg p\)**: This column is the negation of \(p\).
5. **Column for \(\neg p \land q\)**: This column is true only if both \(\neg p\) is true and \(q\) is true.
6. **Final Column**: This column combines the results of \(\neg(p \lor q)\) and \(\neg p \land q\) using the logical OR operator.

### Step 3: Analyze the Final Column
The final column represents the expression \(- (p \lor q) \lor (\neg p \land q)\). We compare this column with the column for \(\neg p\).

### Conclusion
From the truth table, we see that the values in the final column match the values in the \(\neg p\) column for all combinations of \(p\) and \(q\). Therefore, we conclude that:

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