Question 1208947: Prove that the argument is true using the method of natural deduction. You may use cp or ip.
1. (N wedge R) horseshoe B
2. A wedge tilde(M wedge N) / therefore A horseshoe B
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! Therefore, A --> B
This derivation uses the following rules of inference:
Given: Premises provided in the argument.
Assumption for CP: Temporarily assume a statement for the purpose of Conditional Proof.
Reiteration: Repeat a previously derived line.
Disjunctive Syllogism: If P v Q and ~P, then Q.
De Morgan's Law: Rules for negating conjunctions and disjunctions.
Commutation: Change the order of propositions in a disjunction or conjunction.
Material Implication: Equivalence between implications and disjunctions (e.g., P --> Q ≡ ~P v Q).
Exportation: (P & Q) --> R ≡ P --> (Q --> R)
Conjunction Introduction: If P and Q are true, then P & Q is true.
Material Implication (reverse): ~P v Q ≡ P --> Q
Transposition: P --> Q ≡ ~Q --> ~P
Hypothetical Syllogism: If P --> Q and Q --> R, then P --> R.
Conditional Proof (CP): If you can derive Q by assuming P, then you can conclude P --> Q.
This derivation demonstrates that the argument is valid by showing that the conclusion (A --> B) logically follows from the given premises.1. (N & R) → B (Given)
2. A ∨ ~(M & N) (Given)
3. | A (Assumption for Conditional Proof)
4. | A ∨ ~(M & N) (Reiteration 3)
5. | ~(M & N) (Disjunctive Syllogism 2, 4)
6. | ~M ∨ ~N (De Morgan's Law 5)
7. | N → ~M (Material Implication 6)
8. | (N & R) → B (Reiteration 1)
9. | N → (R → B) (Exportation 8)
10. | N → (~M & (R → B)) (Conjunction Introduction 7, 9)
11. | ~(~M & (R → B)) → ~N (Material Implication 10)
12. | (M ∨ ~R) ∨ ~B (De Morgan's Law 11)
13. | ~N ∨ (M ∨ ~R) (Commutation 12)
14. | (N → M) ∨ ~R (Material Implication 13)
15. | (N → M) ∨ (N → ~B) (Material Implication 14)
16. | ~(N → M) → (N → ~B) (Material Implication 15)
17. | A → ~(N → M) (Material Implication 5)
18. | A → (N → ~B) (Hypothetical Syllogism 16, 17)
19. | A → ~B (Exportation 18)
20. A → B (Conditional Proof 3-19)
Therefore, A → B
This derivation demonstrates that the argument is valid using the method of natural deduction and the specified rules of inference.
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