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Logic Proof: Prove r from p→q, ¬p→r, (s∧q)→w, ¬w, s
Mathematics
High School
Question Content
Given: \( p \rightarrow q \), \( \neg p \rightarrow r \), \( (s \land q) \rightarrow w \), \( \neg w \), \( s \). Prove: \( r \).
Correct Answer
r is proven (via logical inference steps)
Detailed Solution Steps
1
1. Apply Modus Tollens (MT) to \( (s \land q) \rightarrow w \) and \( \neg w \): conclude \( \neg (s \land q) \).
2
2. Simplify \( \neg (s \land q) \) using De Morgan’s Law: \( \neg s \lor \neg q \).
3
3. Since \( s \) is true, \( \neg s \) is false. By Disjunctive Syllogism (DS) on \( \neg s \lor \neg q \), conclude \( \neg q \).
4
4. Apply Modus Tollens (MT) to \( p \rightarrow q \) and \( \neg q \): conclude \( \neg p \).
5
5. Apply Modus Ponens (MP) to \( \neg p \rightarrow r \) and \( \neg p \): conclude \( r \).
Knowledge Points Involved
1
Modus Tollens (MT)
Inference rule: If \( A \rightarrow B \) is true and \( \neg B \) is true, then \( \neg A \) is true. Applied to \( (s \land q) \rightarrow w \) (with \( \neg w \)) and \( p \rightarrow q \) (with \( \neg q \)) to derive \( \neg (s \land q) \) and \( \neg p \).
2
De Morgan’s Laws
Used to simplify \( \neg (s \land q) \) to \( \neg s \lor \neg q \), breaking down the negated conjunction.
3
Disjunctive Syllogism (DS)
Applied to \( \neg s \lor \neg q \) and \( s \) to derive \( \neg q \), eliminating the false disjunct \( \neg s \).
4
Modus Ponens (MP)
Applied to \( \neg p \rightarrow r \) (with \( \neg p \)) to derive \( r \), following the implication rule.
5
Logical Implication (\( \rightarrow \))
Used in premises \( p \rightarrow q \), \( \neg p \rightarrow r \), and \( (s \land q) \rightarrow w \) to derive conclusions via MT and MP.
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