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Logic Proof: Prove T from Q→R, R→T, S, ¬(S∧¬Q)
Mathematics
High School
Question Content
Given: \( Q \rightarrow R \), \( R \rightarrow T \), \( S \), \( \neg (S \land \neg Q) \). Prove: \( T \).
Correct Answer
T is proven (via logical inference steps)
Detailed Solution Steps
1
1. Simplify \( \neg (S \land \neg Q) \) using De Morgan’s Law: \( \neg S \lor Q \).
2
2. Since \( S \) is true, \( \neg S \) is false. By Disjunctive Syllogism (DS) on \( \neg S \lor Q \), conclude \( Q \).
3
3. Apply Modus Ponens (MP) to \( Q \rightarrow R \) (with \( Q \)) to get \( R \).
4
4. Apply Modus Ponens (MP) to \( R \rightarrow T \) (with \( R \)) to conclude \( T \).
Knowledge Points Involved
1
De Morgan’s Laws
For logical connectives: \( \neg (A \land B) \equiv \neg A \lor \neg B \) and \( \neg (A \lor B) \equiv \neg A \land \neg B \). Used to simplify \( \neg (S \land \neg Q) \) to \( \neg S \lor Q \).
2
Disjunctive Syllogism (DS)
Inference rule: If \( \neg A \lor B \) is true and \( A \) is true, then \( B \) is true. Applied to \( \neg S \lor Q \) (with \( S \)) to derive \( Q \).
3
Modus Ponens (MP)
Inference rule: If \( A \rightarrow B \) is true and \( A \) is true, then \( B \) is true. Applied to \( Q \rightarrow R \) (with \( Q \)) and \( R \rightarrow T \) (with \( R \)) to derive \( R \) and \( T \).
4
Logical Implication (\( \rightarrow \))
A connective where \( A \rightarrow B \) is true unless \( A \) is true and \( B \) is false. Used to chain implications \( Q \rightarrow R \) and \( R \rightarrow T \) to derive \( T \).
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