Construct the truth table for`[(~p)^^q]implies(pvvq)`

Answer

To construct the truth table for the expression \([(~p) \land q] \implies (p \lor q)\), we will follow these steps:

### Step 1: Identify the variables
We have two variables: \(p\) and \(q\). Each variable can take values of either 0 (False) or 1 (True).

### Step 2: Create the truth table structure
We will create a table with columns for \(p\), \(q\), \(\sim p\) (negation of \(p\)), \((\sim p) \land q\) (negation of \(p\) conjunction \(q\)), \(p \lor q\) (disjunction of \(p\) and \(q\)), and finally \([(~p) \land q] \implies (p \lor q)\).

### Step 3: Fill in the truth values
We will evaluate the truth values for each column based on the values of \(p\) and \(q\).

| \(p\) | \(q\) | \(\sim p\) | \((\sim p) \land q\) | \(p \lor q\) | \([(~p) \land q] \implies (p \lor q)\) |
|-------|-------|-------------|-----------------------|---------------|-----------------------------------------|
| 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 |

### Step 4: Explanation of each column
1. **Negation of \(p\) (\(\sim p\))**: This is simply the opposite of \(p\).
2. **Conjunction \((\sim p) \land q\)**: This is true (1) only if both \(\sim p\) and \(q\) are true.
3. **Disjunction \(p \lor q\)**: This is true (1) if at least one of \(p\) or \(q\) is true.
4. **Implication \([(~p) \land q] \implies (p \lor q)\)**: This is false (0) only when the first part \((\sim p) \land q\) is true and the second part \(p \lor q\) is false. In all other cases, it is true (1).

### Final Truth Table
The final truth table shows that the implication is always true for all combinations of \(p\) and \(q\).