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.