Can you present the derivation of the following logical statement?
((Q ∧ R) → P), (¬(R → S) → (¬W → Q)), (Q ∨ ¬S) ⊢ (R → (P ∨ W))
To derive the statement (R → (P ∨ W)) from the premises ((Q ∧ R) → P), (¬(R → S) → (¬W → Q)), and (Q ∨ ¬S), we can use a proof by contradiction. We will assume the opposite of what we are trying to prove (i.e., assume ¬(R → (P ∨ W))), and derive a contradiction.

Here's the derivation:

1. ¬(R → (P ∨ W)) (Assumption)
2. ¬(¬R ∨ (P ∨ W)) (Implication rule, equivalent to R ∧ ¬(P ∨ W))
3. R (Disjunctive syllogism, from 2 and ¬(P ∨ W))
4. ¬(P ∨ W) (From 2 and R)
5. ¬P ∧ ¬W (De Morgan's law, from 4)
6. Q ∨ ¬S (Premise)
7. (Q ∧ R) → P (Premise)
8. ¬(R → S) → (¬W → Q) (Premise)
9. ¬(R → S) (Assumption for a proof by contradiction)
10. ¬¬W → Q (From 8 and 9, by contraposition)
11. W ∨ Q (From 6 and 10, by disjunctive syllogism)
12. P ∨ W (From 7, 3, and modus ponens)
13. P (From 5 and 12, by disjunctive syllogism)
14. P ∨ W (From 13, by addition)
15. ¬P ∧ ¬W (From 5)
16. Contradiction: P ∧ ¬P (From 14 and 15)
17. Therefore, (R → (P ∨ W)) is true by contradiction (from 1).

Thus, we have shown that if ((Q ∧ R) → P), (¬(R → S) → (¬W → Q)), and (Q ∨ ¬S) are true, then (R → (P ∨ W)) must also be true.