Natural Deduction in Propositional Logic Help?

Question

Use the first four rules of inference to derive the conclusions of the following symbolized arguments. Please Help! Thanks

(4) 1. P ⊃ (G ⊃ T) 

2. Q ⊃ (T ⊃ E)

3. P

4. Q / G ⊃ E 

(5) 1. ∼W ⊃ [∼W ⊃ (X ⊃ W)] 

2. ∼W / ∼X 

(6) 1. J ⊃ (K ⊃ L) 

2. L ∨ J 

3. ∼L / ∼K 

(7) 1. ∼S ⊃ D 

2. ∼S ∨ (∼D ⊃ K) 

3. ∼D / K 

(8) 1. A ⊃ (E ⊃ ∼F) 

2. H ∨ (∼F ⊃ M) 

3. A 

4. ∼H / E ⊃ M 

(9) 1. ∼G ⊃ (G ∨ ∼A) 

2. ∼A ⊃ (C ⊃ A) 

3. ∼G / ∼C 

(10) 1. N ⊃ (J ⊃ P) 

2. (J ⊃ P) ⊃ (N ⊃ J) 

3. N / P 

 (17) 1. H ∨ (Q ∨ F) 

2. R ∨ (Q ⊃ R) 

3. R ∨ ∼H 

4. ∼R / F 

(18) 1. ∼A ⊃ (B ⊃ ∼C) 

2. ∼D ⊃ (∼C ⊃ A) 

3. D ∨ ∼ A 

4. ∼D / ∼B

Answer 1

(4)

if Q then T-->E and G -->T since P is given.

So if Q then G-->T--->E

if Q then G-->E

Therefore, Q or not(G-->E)

(5) If ~W then ~W --> (X -->W)

Which forces X-->W

By contrapositive, ~X must hold..

Therefore, ~W --> ~X

(6) if ~L then J must hold.

then K-->L

then ~K

(7) since ~D, then by 1, S must hold

By 2, ~D --> K, so K since ~D

(8) Since ~H, ~F-->M;

since A, E-->~F

therefore, E-->~F-->M

(9) Since ~G, G or ~A follows by (1)

Since ~G, ~A must hold;

then by 2 , C-->A

But ~A, so ~C by contrapositive

(10) Since N, J-->P must hold;

By 2, N-->J;

But N holds, so must J.

Therefore P

(17) Since ~R, then ~H and Q-->R

By contrapositive, ~Q holds;

Therefore, by 1, F

(18)

Since ~D, ~C-->A by 2 and

~A by 3...

Therefore by contrapositive, C must hold

therefore, By contrapositive of 1, ~B