Construct truth table for `~[p^^(~q)]` and find which implication has same truth value.

Answer

To construct the truth table for the expression `~[p ∧ (~q)]`, we will follow these steps:

### Step 1: Identify the Variables
We have two variables, \( p \) and \( q \).

### Step 2: Create the Truth Values for \( p \) and \( q \)
The possible truth values for \( p \) and \( q \) are:
- True (T)
- False (F)

### Step 3: List All Combinations of Truth Values
We will create a table with all combinations of truth values for \( p \) and \( q \).

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

### Step 4: Calculate \( \neg q \)
Next, we will calculate the negation of \( q \) (denoted as \( \neg q \)).

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

### Step 5: Calculate \( p \land \neg q \)
Now, we will compute the conjunction \( p \land \neg q \).

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

### Step 6: Calculate \( \neg [p \land \neg q] \)
Finally, we will compute the negation of \( p \land \neg q \).

| \( p \) | \( q \) | \( \neg q \) | \( p \land \neg q \) | \( \neg [p \land \neg q] \) |
|---------|---------|---------------|-----------------------|------------------------------|
| T | T | F | F | T |
| T | F | T | T | F |
| F | T | F | F | T |
| F | F | T | F | T |

### Final Truth Table
The final truth table for the expression `~[p ∧ (~q)]` is:

| \( p \) | \( q \) | \( \neg q \) | \( p \land \neg q \) | \( \neg [p \land \neg q] \) |
|---------|---------|---------------|-----------------------|------------------------------|
| T | T | F | F | T |
| T | F | T | T | F |
| F | T | F | F | T |
| F | F | T | F | T |

### Step 7: Find Implications with Same Truth Value
Now, we need to find which implications have the same truth values as \( \neg [p \land \neg q] \).

From the final column, we see that \( \neg [p \land \neg q] \) is true for the combinations:
- \( (T, T) \)
- \( (F, T) \)
- \( (F, F) \)

And it is false for \( (T, F) \).