1. **State the problem:** We want to show that the compound proposition $ (p \lor q) \land (\neg p \lor q) \land (p \lor \neg q) \land (\neg p \lor \neg q) $ is not satisfiable using resolution. 2. **Recall resolution rule:** Resolution is a rule of inference that allows us to derive a new clause from two clauses containing complementary literals. If we derive the empty clause, it means the original set of clauses is unsatisfiable. 3. **Express the proposition as clauses:** - Clause 1: $p \lor q$ - Clause 2: $\neg p \lor q$ - Clause 3: $p \lor \neg q$ - Clause 4: $\neg p \lor \neg q$ 4. **Apply resolution step-by-step:** - Resolve Clause 1 ($p \lor q$) and Clause 4 ($\neg p \lor \neg q$) on $p$: $$ (p \lor q), (\neg p \lor \neg q) \Rightarrow q \lor \neg q $$ This simplifies to a tautology, so no new information. - Resolve Clause 2 ($\neg p \lor q$) and Clause 3 ($p \lor \neg q$) on $p$: $$ (\neg p \lor q), (p \lor \neg q) \Rightarrow q \lor \neg q $$ Again, a tautology. - Resolve Clause 1 ($p \lor q$) and Clause 2 ($\neg p \lor q$) on $p$: $$ (p \lor q), (\neg p \lor q) \Rightarrow q $$ - Resolve Clause 3 ($p \lor \neg q$) and Clause 4 ($\neg p \lor \neg q$) on $p$: $$ (p \lor \neg q), (\neg p \lor \neg q) \Rightarrow \neg q $$ 5. **Derive contradiction:** - From previous steps, we have derived $q$ and $\neg q$. - Resolving $q$ and $\neg q$ gives the empty clause $\square$. 6. **Conclusion:** The empty clause means the original set of clauses is unsatisfiable. Therefore, the compound proposition is not satisfiable.