Math Problem Statement
Prove that the following argument is valid: p →∼ q, r → q, r `∼ p.
Solution
To prove that the argument is valid, we will use natural deduction to show that the conclusion follows logically from the premises.
Given premises:
- (Premise)
- (Premise)
- (Premise)
To prove:
(i.e., the negation of ).
Proof:
- (Premise 1)
- (Premise 2)
- (Premise 3)
- From 2 and 3, by Modus Ponens: (If and is true, then must be true).
- (Negation of ) must be true because of premise 1 (if were true, then must be false, since ).
- Assume is true temporarily (Assumption for Indirect Proof or Reductio ad Absurdum).
- From , by premise 1 (), we get (from 1 and 6).
- But, from step 4, we already know that is true.
- This leads to a contradiction because and cannot both be true.
- Therefore, the assumption that is true must be false.
- Hence, (by Negation Introduction, or Reductio ad Absurdum).
Conclusion:
The conclusion follows from the premises. Therefore, the argument is valid.
Do you want more details or have any questions about this proof?
Here are 5 related questions to explore:
- What is the Modus Ponens rule in logic?
- How does Reductio ad Absurdum work in proving logical statements?
- Can we use a truth table to verify this argument’s validity?
- What is the role of assumptions in natural deduction proofs?
- How would the proof change if were false?
Tip: In proofs by contradiction, always aim to show that assuming the negation of the desired conclusion leads to a contradiction.
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Math Problem Analysis
Mathematical Concepts
Propositional Logic
Natural Deduction
Modus Ponens
Reductio ad Absurdum
Formulas
Modus Ponens: If p → q and p are both true, then q must be true.
Negation Introduction: If assuming p leads to a contradiction, then ∼ p is true.
Theorems
Modus Ponens
Negation Introduction (Reductio ad Absurdum)
Suitable Grade Level
College-level logic or advanced high school
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