To prove that the expression \((p \land q) \land \neg (p \lor q)\) is a contradiction, we will construct a truth table step by step.
### Step 1: Identify the Variables
We have two variables: \(p\) and \(q\).
### Step 2: Create the Truth Table Structure
We will create a truth table with the following columns:
1. \(p\)
2. \(q\)
3. \(p \lor q\) (disjunction)
4. \(\neg (p \lor q)\) (negation of disjunction)
5. \(p \land q\) (conjunction)
6. \((p \land q) \land \neg (p \lor q)\) (final expression)
### Step 3: Fill in the Values for \(p\) and \(q\)
Since there are two variables, \(p\) and \(q\), we will have \(2^2 = 4\) combinations of truth values:
- \(p = T, q = T\)
- \(p = T, q = F\)
- \(p = F, q = T\)
- \(p = F, q = F\)
### Step 4: Calculate \(p \lor q\)
Now we will calculate the values for \(p \lor q\):
- If \(p = T\) and \(q = T\), then \(p \lor q = T\)
- If \(p = T\) and \(q = F\), then \(p \lor q = T\)
- If \(p = F\) and \(q = T\), then \(p \lor q = T\)
- If \(p = F\) and \(q = F\), then \(p \lor q = F\)
### Step 5: Calculate \(\neg (p \lor q)\)
Next, we find the negation of \(p \lor q\):
- If \(p \lor q = T\), then \(\neg (p \lor q) = F\)
- If \(p \lor q = F\), then \(\neg (p \lor q) = T\)
### Step 6: Calculate \(p \land q\)
Now we calculate \(p \land q\):
- If \(p = T\) and \(q = T\), then \(p \land q = T\)
- If \(p = T\) and \(q = F\), then \(p \land q = F\)
- If \(p = F\) and \(q = T\), then \(p \land q = F\)
- If \(p = F\) and \(q = F\), then \(p \land q = F\)
### Step 7: Calculate \((p \land q) \land \neg (p \lor q)\)
Finally, we calculate the final expression:
- If \(p \land q = T\) and \(\neg (p \lor q) = F\), then \((p \land q) \land \neg (p \lor q) = F\)
- If \(p \land q = F\) and \(\neg (p \lor q) = F\), then \((p \land q) \land \neg (p \lor q) = F\)
- If \(p \land q = F\) and \(\neg (p \lor q) = F\), then \((p \land q) \land \neg (p \lor q) = F\)
- If \(p \land q = F\) and \(\neg (p \lor q) = T\), then \((p \land q) \land \neg (p \lor q) = F\)
### Step 8: Compile the Truth Table
Now we can compile the truth table:
| \(p\) | \(q\) | \(p \lor q\) | \(\neg (p \lor q)\) | \(p \land q\) | \((p \land q) \land \neg (p \lor q)\) |
|-------|-------|---------------|----------------------|----------------|---------------------------------------|
| T | T | T | F | T | F |
| T | F | T | F | F | F |
| F | T | T | F | F | F |
| F | F | F | T | F | F |
### Conclusion
The last column of the truth table shows that \((p \land q) \land \neg (p \lor q)\) is always false, which means it is a contradiction (fallacy).
