1. **State the problem:** Verify the logical proposition $((P \wedge Q) \wedge E) \to (P \vee q)$. 2. **Understand the proposition:** The formula is an implication where the antecedent is $((P \wedge Q) \wedge E)$ and the consequent is $(P \vee q)$. 3. **Recall implication property:** An implication $A \to B$ is false only when $A$ is true and $B$ is false. In all other cases, it is true. 4. **Analyze the antecedent:** $((P \wedge Q) \wedge E)$ means all three propositions $P$, $Q$, and $E$ are true. 5. **Analyze the consequent:** $(P \vee q)$ means either $P$ or $q$ (or both) is true. 6. **Check when implication could fail:** The antecedent is true when $P$, $Q$, and $E$ are all true. 7. **If $P$ is true, then $(P \vee q)$ is true regardless of $q$ since $P \vee q$ is true if $P$ is true. 8. Since the antecedent demands $P$ to be true (as part of the conjunction), the consequent $(P \vee q)$ must always be true when the antecedent is true. 9. **Conclusion:** The implication $((P \wedge Q) \wedge E) \to (P \vee q)$ is always true (a tautology). 10. **Final answer:** The proposition is valid because whenever the antecedent is true, the consequent is inevitably true.