Hodges Crawford
01/10/2024 · Elementary School
Construct a truth table for the statement \( \sim(q \wedge r) \rightarrow(\sim p \wedge r) \).
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Step-by-step Solution
To construct a truth table for the statement \( \sim(q \wedge r) \rightarrow(\sim p \wedge r) \), we will first identify the variables involved: \( p \), \( q \), and \( r \). We will evaluate the truth values of the components of the statement step by step.
1. **Identify the components**:
- \( p \)
- \( q \)
- \( r \)
- \( q \wedge r \) (the conjunction of \( q \) and \( r \))
- \( \sim(q \wedge r) \) (the negation of \( q \wedge r \))
- \( \sim p \) (the negation of \( p \))
- \( \sim p \wedge r \) (the conjunction of \( \sim p \) and \( r \))
- \( \sim(q \wedge r) \rightarrow(\sim p \wedge r) \) (the implication)
2. **Create the truth table**:
| \( p \) | \( q \) | \( r \) | \( q \wedge r \) | \( \sim(q \wedge r) \) | \( \sim p \) | \( \sim p \wedge r \) | \( \sim(q \wedge r) \rightarrow(\sim p \wedge r) \) |
|---------|---------|---------|-------------------|-------------------------|---------------|------------------------|-----------------------------------------------------|
| T | T | T | T | F | F | F | T |
| T | T | F | F | T | F | F | F |
| T | F | T | F | T | F | F | F |
| T | F | F | F | T | F | F | F |
| F | T | T | T | F | T | T | T |
| F | T | F | F | T | T | F | F |
| F | F | T | F | T | T | T | T |
| F | F | F | F | T | T | F | F |
3. **Explanation of the columns**:
- The first three columns represent all possible combinations of truth values for \( p \), \( q \), and \( r \).
- The fourth column calculates \( q \wedge r \).
- The fifth column calculates \( \sim(q \wedge r) \).
- The sixth column calculates \( \sim p \).
- The seventh column calculates \( \sim p \wedge r \).
- The last column calculates the implication \( \sim(q \wedge r) \rightarrow(\sim p \wedge r) \).
4. **Final results**:
- The implication \( A \rightarrow B \) is false only when \( A \) is true and \( B \) is false. In all other cases, it is true.
Thus, the truth table for the statement \( \sim(q \wedge r) \rightarrow(\sim p \wedge r) \) is complete.
Quick Answer
The truth table for \( \sim(q \wedge r) \rightarrow(\sim p \wedge r) \) is as follows:
| \( p \) | \( q \) | \( r \) | \( \sim(q \wedge r) \) | \( \sim p \) | \( \sim p \wedge r \) | \( \sim(q \wedge r) \rightarrow(\sim p \wedge r) \) |
|---------|---------|---------|-------------------------|---------------|------------------------|-----------------------------------------------------|
| T | T | T | F | F | F | T |
| T | T | F | T | F | F | F |
| T | F | T | T | F | F | F |
| T | F | F | T | F | F | F |
| F | T | T | F | T | T | T |
| F | T | F | T | T | F | F |
| F | F | T | T | T | T | T |
| F | F | F | T | T | F | F |
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