1. **Problem Statement:** Prove that the compound proposition $((p \to q) \wedge (q \to r)) \to (p \to r)$ is a tautology using algebraic laws of propositional logic. 2. **Recall the implication equivalence:** $p \to q$ is logically equivalent to $\neg p \lor q$. 3. **Rewrite the proposition:** $$((p \to q) \wedge (q \to r)) \to (p \to r)$$ becomes $$((\neg p \lor q) \wedge (\neg q \lor r)) \to (\neg p \lor r)$$ 4. **Rewrite the implication as disjunction:** $$\neg ((\neg p \lor q) \wedge (\neg q \lor r)) \lor (\neg p \lor r)$$ 5. **Apply De Morgan's law to the negation:** $$ (\neg (\neg p \lor q) \lor \neg (\neg q \lor r)) \lor (\neg p \lor r)$$ 6. **Apply De Morgan's law inside:** $$ ((p \wedge \neg q) \lor (q \wedge \neg r)) \lor (\neg p \lor r)$$ 7. **Use associativity and commutativity to regroup:** $$ (p \wedge \neg q) \lor (q \wedge \neg r) \lor \neg p \lor r$$ 8. **Group terms:** $$ (p \wedge \neg q) \lor \neg p \lor (q \wedge \neg r) \lor r$$ 9. **Use absorption and distributive laws:** - $p \wedge \neg q \lor \neg p = (p \lor \neg p) \wedge (\neg q \lor \neg p) = \text{True} \wedge (\neg q \lor \neg p) = \neg q \lor \neg p$ - Similarly, $q \wedge \neg r \lor r = (q \lor r) \wedge (\neg r \lor r) = (q \lor r) \wedge \text{True} = q \lor r$ 10. **Substitute back:** $$ (\neg q \lor \neg p) \lor (q \lor r)$$ 11. **Use commutativity:** $$ \neg p \lor \neg q \lor q \lor r$$ 12. **Apply complement law:** $\neg q \lor q = \text{True}$, so $$ \neg p \lor \text{True} \lor r = \text{True}$$ 13. **Conclusion:** The expression simplifies to True, so the original proposition is a tautology. **Final answer:** The compound proposition $((p \to q) \wedge (q \to r)) \to (p \to r)$ is a tautology.