To solve the Boolean expression \(((p \land q) \lor (p \lor \neg q)) \land (\neg p \land \neg q)\) and find its equivalent, we will simplify the expression step by step.
### Step 1: Break down the expression
The given expression is:
\[
((p \land q) \lor (p \lor \neg q)) \land (\neg p \land \neg q)
\]
We can simplify the expression inside the first parentheses.
### Step 2: Simplify the first part \((p \land q) \lor (p \lor \neg q)\)
We can use the distributive property of Boolean algebra:
1. \(p \lor \neg q\) can be simplified further.
2. The expression \((p \land q) \lor (p \lor \neg q)\) can be evaluated by considering the possible values of \(p\) and \(q\).
### Step 3: Evaluate the truth values
We will create a truth table for \(p\) and \(q\):
| p | q | \(p \land q\) | \(\neg q\) | \(p \lor \neg q\) | \((p \land q) \lor (p \lor \neg q)\) |
|-------|-------|----------------|------------|--------------------|--------------------------------------|
| T | T | T | F | T | T |
| T | F | F | T | T | T |
| F | T | F | F | F | F |
| F | F | F | T | T | T |
From the table, we see that:
\[
(p \land q) \lor (p \lor \neg q) = T \text{ for } (T, T), (T, F), (F, F) \text{ and } F \text{ for } (F, T).
\]
### Step 4: Combine with \((\neg p \land \neg q)\)
Next, we need to evaluate \(\neg p \land \neg q\):
| p | q | \(\neg p\) | \(\neg q\) | \(\neg p \land \neg q\) |
|-------|-------|------------|------------|--------------------------|
| T | T | F | F | F |
| T | F | F | T | F |
| F | T | T | F | F |
| F | F | T | T | T |
### Step 5: Evaluate the final expression
Now we combine the results from the previous steps:
\[
((p \land q) \lor (p \lor \neg q)) \land (\neg p \land \neg q)
\]
From the truth table, we see that \((p \land q) \lor (p \lor \neg q)\) is true for the combinations (T, T), (T, F), and (F, F), but \(\neg p \land \neg q\) is only true for (F, F).
Thus, the final expression evaluates to:
- True when both parts are true, which only happens when \(p\) and \(q\) are both false.
### Conclusion
The equivalent expression for \(((p \land q) \lor (p \lor \neg q)) \land (\neg p \land \neg q)\) is:
\[
\neg p \land \neg q
\]
