Question 1179696
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.