1. **State the problem:** We want to analyze the logical implication $(P \wedge q) \Rightarrow (P \vee \overline{q})$. 2. **Recall the implication rule:** An implication $A \Rightarrow B$ is logically equivalent to $\overline{A} \vee B$. 3. **Apply the rule:** $$(P \wedge q) \Rightarrow (P \vee \overline{q}) \equiv \overline{(P \wedge q)} \vee (P \vee \overline{q})$$ 4. **Use De Morgan's law on $\overline{(P \wedge q)}$:** $$\overline{P} \vee \overline{q}$$ 5. **Substitute back:** $$\overline{P} \vee \overline{q} \vee P \vee \overline{q}$$ 6. **Simplify by idempotent and commutative laws:** $$\overline{P} \vee P \vee \overline{q}$$ 7. **Since $\overline{P} \vee P$ is a tautology (always true), the whole expression is true regardless of $q$:** **Final answer:** The implication $(P \wedge q) \Rightarrow (P \vee \overline{q})$ is a tautology (always true). --- Since the second problem $P \Rightarrow \overline{P}$ is not solved as per instructions, it is counted but not answered.