Construct truth tables for each of the following :
`[(~p)^^ q]implies(p vv q)`

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 \(q\).

### Step 2: Determine the number of rows in the truth table
Since there are 2 variables, the number of combinations of truth values is \(2^2 = 4\). Therefore, we will have 4 rows in our truth table.

### Step 3: Create the initial structure of the truth table
We will create a table with the following columns:
1. \(p\)
2. \(q\)
3. \(\sim p\) (negation of \(p\))
4. \(\sim p \land q\) (conjunction of \(\sim p\) and \(q\))
5. \(p \lor q\) (disjunction of \(p\) and \(q\))
6. \((\sim p \land q) \implies (p \lor q)\) (the final expression)

### Step 4: Fill in the truth values for \(p\) and \(q\)
The possible combinations of truth values for \(p\) and \(q\) are:
- \(T, T\)
- \(T, F\)
- \(F, T\)
- \(F, F\)

### Step 5: Calculate \(\sim p\)
- If \(p\) is \(T\), then \(\sim p\) is \(F\).
- If \(p\) is \(F\), then \(\sim p\) is \(T\).

### Step 6: Calculate \(\sim p \land q\)
- This conjunction is true if both \(\sim p\) and \(q\) are true, otherwise it is false.

### Step 7: Calculate \(p \lor q\)
- This disjunction is true if at least one of \(p\) or \(q\) is true.

### Step 8: Calculate \((\sim p \land q) \implies (p \lor q)\)
- This implication is false only when the antecedent (\(\sim p \land q\)) is true and the consequent (\(p \lor q\)) is false; otherwise, it is true.

### Step 9: Fill in the truth table
Now, we will fill in the truth table based on the calculations:

| \(p\) | \(q\) | \(\sim p\) | \(\sim p \land q\) | \(p \lor q\) | \((\sim p \land q) \implies (p \lor q)\) |
|-------|-------|------------|---------------------|---------------|------------------------------------------|
| T | T | F | F | T | T |
| T | F | F | F | T | T |
| F | T | T | T | T | T |
| F | F | T | F | F | T |

### Final Result
The completed truth table for the expression \([(~p) \land q] \implies (p \lor q)\) is as shown above.