create truth table for ∼(*p*∧*q*)→*p*.

To create a truth table for the expression \(\sim(p \land q) \rightarrow p\), we need to break down the expression step-by-step and evaluate the result for all possible values of \(p\) and \(q\).

The steps are as follows:

1. Determine the values of \(p \land q\).
2. Negate \(p \land q\) to find \(\sim(p \land q)\).
3. Evaluate the implication \(\sim(p \land q) \rightarrow p\).

### Truth Table

| \( p \) | \( q \) | \( p \land q \) | \( \sim(p \land q) \) | \( \sim(p \land q) \rightarrow p \) |
|---------|---------|-----------------|------------------------|------------------------------------|
| T       | T       | T               | F                      | T                                  |
| T       | F       | F               | T                      | T                                  |
| F       | T       | F               | T                      | F                                  |
| F       | F       | F               | T                      | F                                  |

### Explanation of the Results

- When \(p\) and \(q\) are both true, \(p \land q\) is true, so \(\sim(p \land q)\) is false. The implication \(\sim(p \land q) \rightarrow p\) is true because an implication with a false antecedent is true.
- When \(p\) is true and \(q\) is false, \(\sim(p \land q)\) is true, and the implication is true since \(p\) itself is true.
- When \(p\) is false and \(q\) is true, \(\sim(p \land q)\) is true, but the implication is false because \(p\) (the consequent) is false.
- When both \(p\) and \(q\) are false, \(\sim(p \land q)\) is true, but the implication is false as \(p\) is false.

Would you like further details or have questions?

Here are five related questions for exploration:
1. How does the implication \(\rightarrow\) affect truth tables compared to other operators?
2. What would the truth table look like for \(p \rightarrow (p \land q)\)?
3. How does the negation of a conjunction \(\sim(p \land q)\) compare to \(\sim p \lor \sim q\)?
4. How would the truth table change if we used \(\sim(p \lor q) \rightarrow p\) instead?
5. How does the truth value of \(\sim(p \land q)\) affect other logical statements?

**Tip:** In logic, an implication \(A \rightarrow B\) is always true if \(A\) is false, regardless of \(B\)'s value.

Create truth table for ∼(p ∧ q) → p.

Explore how to create a truth table for the logical expression ∼(p ∧ q) → p, breaking down each step in propositional logic. This solution covers fundamental concepts such as negation, conjunction, and implication, including a detailed truth table and explanation. Suitable for high school students in Grades 10-12.

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