Use rules of inference and without truth table, show that ((p ∨ q) ∧ (¬p ∨ r)) → (q ∨r) is a valid argument.

To show that \(((p \lor q) \land (\neg p \lor r)) \to (q \lor r)\) is a valid argument using rules of inference, we'll proceed by applying logical equivalences and inference rules step by step. The goal is to show that the premises logically lead to the conclusion, without the use of a truth table.

### Steps:

1. **Rewrite the implication**:  
   \[
   ((p \lor q) \land (\neg p \lor r)) \to (q \lor r)
   \]
   can be rewritten using the **implication equivalence**:  
   \[
   A \to B \equiv \neg A \lor B
   \]
   This gives us:
   \[
   \neg((p \lor q) \land (\neg p \lor r)) \lor (q \lor r)
   \]
   
2. **Apply De Morgan's Law** to the negation of the conjunction \(\neg((p \lor q) \land (\neg p \lor r))\):  
   According to **De Morgan’s law**, the negation of a conjunction is the disjunction of the negations:
   \[
   \neg((p \lor q) \land (\neg p \lor r)) \equiv \neg(p \lor q) \lor \neg(\neg p \lor r)
   \]
   
   Using **De Morgan's law** again, we simplify further:  
   \[
   \neg(p \lor q) \equiv \neg p \land \neg q \quad \text{and} \quad \neg(\neg p \lor r) \equiv p \land \neg r
   \]
   So, we now have:
   \[
   (\neg p \land \neg q) \lor (p \land \neg r)
   \]
   
3. **Distribute the disjunction**:  
   Now we apply the **distributive property** of disjunction over conjunction:
   \[
   (\neg p \land \neg q) \lor (p \land \neg r)
   \]
   This is already simplified, but it represents two distinct cases:
   - Case 1: \(\neg p \land \neg q\)
   - Case 2: \(p \land \neg r\)
   
4. **Consider both cases**:

   - In **Case 1**: If \(\neg p \land \neg q\) is true, then \(q\) is false, so \(q \lor r\) depends on \(r\). Since \(\neg q\) is true, \(q \lor r\) will be true if \(r\) is true, meaning \(q \lor r\) holds. Hence, the conclusion holds.
   
   - In **Case 2**: If \(p \land \neg r\) is true, then \(p\) is true and \(r\) is false. In this case, \(p\) being true implies that \(p \lor q\) is true, but since \(r\) is false, \(q \lor r\) depends on \(q\). Since we don't have \(\neg q\) in this case, \(q \lor r\) could still hold if \(q\) is true. Thus, the conclusion holds.

5. **Conclusion**:  
   Since in both cases the conclusion \(q \lor r\) holds, the argument is valid.

### Final Statement:
Thus, the argument \(((p \lor q) \land (\neg p \lor r)) \to (q \lor r)\) is valid, as we have shown that the premises lead logically to the conclusion using rules of inference.

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Would you like more details or have any questions about this solution?  
Here are 5 related questions you can explore further:

1. How do De Morgan's laws apply to negations of complex statements?
2. What is the significance of the implication equivalence in logical proofs?
3. How does distribution of disjunction over conjunction work in logic?
4. What other rules of inference could be used in this proof?
5. How would this proof change if we had additional premises?

### Tip:
Always look for key equivalences like De Morgan's laws and implication rules—they can drastically simplify complex logical statements!

Use rules of inference and without truth table, show that ((p ∨ q) ∧ (¬p ∨ r)) → (q ∨ r) is a valid argument.

Learn how to validate the argument ((p ∨ q) ∧ (¬p ∨ r)) → (q ∨ r) using rules of inference and logical equivalences like De Morgan's Law and Implication Equivalence. This step-by-step solution is ideal for undergraduate-level discrete mathematics students.

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