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.