SOLUTION: INSTRUCTIONS: Use natural deduction to derive the conclusion in each problem. Prove this using natural deduction. NOTE: Use * for dot, v for wedge, ~ for tilde, = for triple ba

Question 1179696: INSTRUCTIONS: Use natural deduction to derive the conclusion in each problem.
Prove this using natural deduction.
NOTE: Use * for dot, v for wedge, ~ for tilde, = for triple bar (or copy and paste ≡), and > for horseshoe (or copy and paste ⊃ )

1. N ≡ F
2. ~F v ~N
3. D ⊃ N ~(F v D)

1. (B ⊃ G) • (F ⊃ N)
2. ~(G * N) / ~(B * F)

1. (J • R) ⊃ H
2. (R ⊃ H) ⊃ M
3. ~(P v ~J) / M • ~P

1. (F • H) ⊃ N
2 F v S
3. H / N v S
Please any guidance on these 4 questions I'd greatly appreciate it!
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve these natural deduction proofs step-by-step:
1. Proof:
N ≡ F
~F v ~N
D ⊃ N / ~(F v D)
Proof:
(N ⊃ F) * (F ⊃ N) (1, Equivalence)
N ⊃ F (4, Simplification)
F ⊃ N (4, Simplification)
~N ⊃ ~F (5, Contraposition)
~F ⊃ ~N (6, Contraposition)
~N v ~F (7, Implication)
~N (2, 9, Resolution)
~D (3, 10, Modus Tollens)
~F (8, 10, Modus Ponens)
~F * ~D (11, 12, Conjunction)
~(F v D) (13, De Morgan's)
Therefore, ~(F v D) is proven.
2. Proof:
(B ⊃ G) * (F ⊃ N)
~(G * N) / ~(B * F)
Proof:
B ⊃ G (1, Simplification)
F ⊃ N (1, Simplification)
~G v ~N (2, De Morgan's)
~B v G (3, Implication)
~F v N (4, Implication)
~B v ~N (5, 6, Resolution)
~F v ~G (5, 7, Resolution)
~B (8, 9, Resolution)
~F (8, 10, Resolution)
~B * ~F (10, 11, Conjunction)
~(B * F) (12, De Morgan's)
Therefore, ~(B * F) is proven.
3. Proof:
(J * R) ⊃ H
(R ⊃ H) ⊃ M
~(P v ~J) / M * ~P
Proof:
~P * ~~J (3, De Morgan's)
~P (4, Simplification)
~~J (4, Simplification)
J (6, Double Negation)
~(J * ~R) (1, Implication)
~J v ~R (8, De Morgan's)
~R (7, 9, Disjunctive Syllogism)
R ⊃ H (1, Exportation)
M (2, 11, Modus Ponens)
M * ~P (5, 12, Conjunction)
Therefore, M * ~P is proven.
4. Proof:
(F * H) ⊃ N
F v S
H / N v S
Proof:
~F v N (1, Exportation)
F (2, Assumption)
N (4, 5, Disjunctive Syllogism)
N v S (6, Addition)
S (2, Assumption)
N v S (8, Addition)
H (3, Copy)
F * H (5, 10, Conjunction)
N (1, 11, Modus Ponens)
N v S (12, Addition)
N v S (5-7, 8-9, 13, Conditional Proof)
Therefore, N v S is proven.
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