SOLUTION: I need help constructing an indirect proof using reductio ad absurdum for: ~S → (F → L), F → (L → P), therefore, ~S → (F → P) Can the proof be performed more effi

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Question 1150113: I need help constructing an indirect proof using reductio ad absurdum for:
~S → (F → L), F → (L → P), therefore, ~S → (F → P)
Can the proof be performed more efficiently by using different equivalence rules?
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
1.  ~S → (F → L), 
2.  F → (L → P),          therefore, ~S → (F → P)

                  | 3. ~[~S → (F → P)]     Assumption for Indirect Proof
                  | 4. ~[~S → (~F v P)]    3, Material Implication   
                  | 5. ~[~~S v (~F v P)]   4, Material Implication   
                  | 6. ~{S v (~F v P)]     5, Double Negation
                  | 7. ~S & ~(~F v P)      6, DeMorgan's Law
                  | 8. ~S & (~~F & ~P)     7, DeMorgan's Law
                  | 9. ~S & (F & ~P)       8, Double Negation  
                  |10. ~S                  9, Simplification
                  |11. F → L               1,10, Modus Ponens
                  |12. (F & ~P) & ~S       9. Commutation
                  |13. F & (~P & ~S)       12, Association
                  |14. F                   13, Simplification
                  |15. L → P               2,14 Modus Ponens
                  |16. L                   11,14 Modus Ponens
                  |17. P                   15,16 Modus Ponens 
                  |18. (F & ~P) & ~S       9, Commutation
                  |19. (~P & F) & ~S       18, Commutation
                  |20. ~P & (F & ~S)       19, Asociation
                  |21. ~P                  20, Simplification
                  |22. P & ~P              17,21 Conjunction   
23.  ~S → (F → P)   Lines 3-22  Indirect Proof

Edwin