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Proof: Repository - hw4.5
Construct a proof for the argument: ¬(Q → R) → P, ¬P, Q ∴ R

1	¬(Q → R) → P		
2	¬P		
3	Q		
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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 construct a proof for the argument ¬(Q → R) → P, ¬P, Q ∴ R using the formal rules of logic, we can proceed step-by-step as follows. We need to prove \( R \) from the premises:

1. \( \neg(Q \to R) \to P \) — Premise
2. \( \neg P \) — Premise
3. \( Q \) — Premise
∴ \( R \)

### Strategy
We will use **modus tollens (MT)** and **reductio ad absurdum (RAA)** to prove \( R \). Since we are given \( \neg P \), the strategy is to show that assuming \( \neg R \) leads to a contradiction, thereby deriving \( R \).

### Proof Steps:

| Step | Statement               | Justification          |
|------|-------------------------|------------------------|
| 1    | \( \neg(Q \to R) \to P \) | Premise                |
| 2    | \( \neg P \)              | Premise                |
| 3    | \( Q \)                   | Premise                |
| 4    | \( \neg P \)              | Repetition (from 2)    |
| 5    | \( \neg \neg (Q \to R) \) | Modus Tollens (1, 4)   |
| 6    | \( Q \to R \)             | Double Negation Elim (5) |
| 7    | \( R \)                   | Modus Ponens (6, 3)    |

### Explanation:
- Step 1 to 3 sets up the premises of the argument.
- Step 4 repeats the negation of \( P \) for later use.
- Step 5 uses **modus tollens** to infer \( \neg \neg (Q \to R) \) from lines 1 and 2.
- Step 6 applies **double negation elimination** to simplify \( \neg \neg (Q \to R) \) to \( Q \to R \).
- Step 7 applies **modus ponens** using \( Q \) from line 3 and \( Q \to R \) from line 6 to derive \( R \).

Thus, the conclusion \( R \) follows logically from the premises.

Would you like any clarifications or further details?

### Related Questions:
1. What is the difference between modus tollens and modus ponens?
2. How does double negation work in propositional logic?
3. Can reductio ad absurdum be used in this proof as an alternative approach?
4. What is the role of premises in formal proofs?
5. How would the proof change if we were given \( P \) instead of \( \neg P \)?

### Tip:
When working with formal proofs, it's useful to outline your strategy before diving into the detailed steps to ensure you follow a logical path.

Construct a proof for the argument: ¬(Q → R) → P, ¬P, Q ∴ R

Learn how to construct a formal proof for the argument ¬(Q → R) → P, ¬P, Q ∴ R using key inference rules like Modus Ponens, Modus Tollens, and Double Negation. This problem is suitable for undergraduate students studying logic and philosophy.

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