Construct a proof for the argument: P v Q, P -> ~R, R ∴ Q

1P ∨ Q

2P → ¬R

3R

**Notation for logic operators**

negation:

∼

conjunction:

^

disjunction:

v

conditional:

->

biconditional:

<->

universal quantification:

Ax or (Ax)

existential quantification:

Ex or (Ex)

**Rule names (full and abbreviated)**

modus ponens

->E

modus tollens

MT

modus tollendo ponens

DS

double negation

DNE

addition

vI

adjunction

^I

simplification

^E

bicondition

<->I

equivalence

<->E

repeat

Rep

conditional derivation

->I

reductio ad absurdum

RAA

universal instantiation

AE

universal derivation

AI

existential instantiation

EE

existential generalization

EI

identity introduction

=I

substitution of identicals

=E


To prove the argument:

**Premises:**

1. \( P \lor Q \)
2. \( P \rightarrow \neg R \)
3. \( R \)

**Conclusion:**

\( Q \)

We will construct a formal proof using the rules of inference, as outlined.

---

### Proof:

1. \( P \lor Q \)  — **Premise**  
2. \( P \rightarrow \neg R \)  — **Premise**  
3. \( R \)  — **Premise**  
4. \( \neg \neg R \) — **Double Negation Introduction (DNE)** on (3)  
5. \( \neg P \) — **Modus Tollens (MT)** using (2) and (4)  
6. \( Q \) — **Disjunction Elimination (DS)** on (1) and (5)

### Explanation:

1. The first step assumes the disjunction \( P \lor Q \).
2. The second step introduces the conditional \( P \rightarrow \neg R \).
3. The third step introduces the premise \( R \).
4. From \( R \), we introduce double negation \( \neg \neg R \), allowing the use of Modus Tollens.
5. Using Modus Tollens on \( P \rightarrow \neg R \) and \( \neg \neg R \), we derive \( \neg P \).
6. Finally, using Disjunction Elimination (Modus Tollendo Ponens, DS) on \( P \lor Q \) and \( \neg P \), we conclude \( Q \).

Thus, \( Q \) is proven.

---

Do you have any questions or want further details on any step?

Here are five related questions to explore further:

1. How does Modus Tollens work with conditionals in logical proofs?
2. Can \( P \lor Q \) be true if both \( P \) and \( Q \) are false?
3. How does Disjunction Elimination (DS) help simplify logical arguments?
4. What is the importance of double negation in formal logic?
5. Can we reach the same conclusion if the premises were altered slightly?

**Tip:** Modus Tollens is often useful when the negation of the conclusion helps derive the negation of the premise.

Construct a proof for the argument: P v Q, P -> ~R, R ∴ Q

Explore the step-by-step construction of a logical proof for the argument P ∨ Q, P -> ~R, R ∴ Q. This problem involves key concepts in propositional logic, including Modus Tollens, Disjunction Elimination, and Double Negation. Suitable for undergraduate students studying logic.

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