Prove that the following argument is valid: p →∼ q, r → q, r `∼ p.

To prove that the argument \( p \rightarrow \sim q, \ r \rightarrow q, \ r \ \vdash \ \sim p \) is valid, we will use **natural deduction** to show that the conclusion \( \sim p \) follows logically from the premises.

### Given premises:
1. \( p \rightarrow \sim q \)   (Premise)
2. \( r \rightarrow q \)        (Premise)
3. \( r \)                      (Premise)

### To prove:
\( \sim p \) (i.e., the negation of \( p \)).

### Proof:
1. \( p \rightarrow \sim q \)   (Premise 1)
2. \( r \rightarrow q \)        (Premise 2)
3. \( r \)                      (Premise 3)
4. From 2 and 3, by **Modus Ponens**: \( q \)      (If \( r \rightarrow q \) and \( r \) is true, then \( q \) must be true).
5. \( \sim q \) (Negation of \( q \)) must be true because of premise 1 (if \( p \) were true, then \( q \) must be false, since \( p \rightarrow \sim q \)).
6. Assume \( p \) is true temporarily (Assumption for **Indirect Proof** or **Reductio ad Absurdum**).
7. From \( p \), by premise 1 (\( p \rightarrow \sim q \)), we get \( \sim q \) (from 1 and 6).
8. But, from step 4, we already know that \( q \) is true.
9. This leads to a **contradiction** because \( q \) and \( \sim q \) cannot both be true.
10. Therefore, the assumption that \( p \) is true must be false.
11. Hence, \( \sim p \) (by **Negation Introduction**, or **Reductio ad Absurdum**).

### Conclusion:
The conclusion \( \sim p \) follows from the premises. Therefore, the argument is **valid**.

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Do you want more details or have any questions about this proof?

Here are 5 related questions to explore:
1. What is the **Modus Ponens** rule in logic?
2. How does **Reductio ad Absurdum** work in proving logical statements?
3. Can we use a truth table to verify this argument’s validity?
4. What is the role of assumptions in natural deduction proofs?
5. How would the proof change if \( r \) were false?

**Tip**: In proofs by contradiction, always aim to show that assuming the negation of the desired conclusion leads to a contradiction.

Prove that the following argument is valid: p →∼ q, r → q, r ∴ ∼ p.

Discover how to prove the validity of the argument p → ∼q, r → q, r ∴ ∼p using natural deduction. This solution involves key concepts like Modus Ponens and Reductio ad Absurdum, suitable for college-level logic or advanced high school students.

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