philosophy logic (Natural Deduction in Propositional Logic )

Question

If high-tech products are exported to Russia, then domestic industries will benefit. If the Russians can effectively utilize high-tech products, then their standard of living will improve. So, if high-tech products are exported to Russia and the Russians can effectively utilize them, then their standard of living will improve and domestic industries will benefit.

(H, D, U, S)

P.S: indirect proof(IP) cannot be used! 

- suggest using Reductio ad absurdum and Conditional Proof if needed 

- Solve the problem in a different way than the example answer below...

1. H --> D

2. U --> S                  So (H•U) --> (S•D) 

3. | H•U                   Assume CP

4. | H                        3, Simp

5. | D                        1,4, MP

6. | U•H                     3, Com

7. | U                         6, Simp

8. | S                        2,7, MP

9. | S•D                     5,8, Conj

10. (H•U) --> (S•D)       3-9, CP

Answer 1

1.     H > D

2.     U > S

3.     ~U v (H > D)    1 Add.

4.     ~U v (~H v D)   3 Impl.

5.     (~U v ~H) v D   4 Asso.

6.     ~(U & H) v D   5 DM

7.     ~(H & U) v D   6 Comm.

8.     ~H v (U > S)    2 Add.

9.     ~H v (~U v S)  8 Impl.

10.  (~H v ~U) v S   9 Asso.

11.   ~(H & U) v S      10 DM

12.  [~(H & U) v S] & [~(H & U) v D] 7,11 Conj.

13.  ~(H & U) v (S & D)                     12 Dist.

14.  (H & U) > (S & D)                       13 Impl.

You can complete this proof without CP or IP. I find the easiest way is to reverse engineer it. Start with the conclusion and work your way back. Just playing around with it, I was able to get " [~(H & U) v S] & [~(H & U) v D] " from the conclusion, after that it was just a matter of figuring out how to get both of those conjuncts from the premises which is fairly straightforward through Addition and Implication. Let me know if you have any questions on this proof!

Answer 2

H-->D is given

U-->S is given

By implication identity:

H-->d = not H or D

U-->s = not U or S

Given all premesis hold true:

(H and U) and [(not H or D)] and [(not U) or S]

Commutative property:

H and [(not H or D)] and U and [(not U) or S]

Distributive:

{(H and not H)or (H and D)} and {(U and not U) or ( U as S)}

Dichotomy:

{False or (H and D) } and {False or (U and S)}

Disjunction:

(H and D) and (U and S)

H-->D and U-->S:

(D and D) and (S and S)

Therefore:

D and S