SOLUTION: ∼F → ∼G, P → ∼Q, ∼F ∨ P, (∼G ∨ ∼Q) → (L • M) ∴ L

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Question 1141256: ∼F → ∼G, P → ∼Q, ∼F ∨ P, (∼G ∨ ∼Q) → (L • M) ∴ L
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
 1.  ∼F → ∼G
 2.  P → ∼Q
 3.  ∼F ∨ P
 4. (∼G ∨ ∼Q) → (L • M)      ∴ L

                  | 5.  ~L            Assumption for indirect proof
                  | 6.  ~L ∨ ~M       5, addition
                  | 7.  ~(L • M)      6, deMorgan's law
                  | 8.  ~(~G ∨ ~Q)    4,7, modus tollens
                  | 9.  ~~G • ~~Q     8, deMorgan's law
                  |10.  G • Q         9, double negation
                  |11.  Q • G         10, commutation
                  |12.  Q             11, simplification
                  |13.  ~~Q           12, double negation
                  |14. ~P             2,13, modus tollens
                  |15. P ∨ ~F         3, commutation
                  |16. ~F             15,14, disjunctive syllogism
                  |17. ~G             1,16, modus ponens
                  |18. G              10, simplification
                  |19. G • ~G         18,17, conjunction
 20. L         lines 5-19  indirect proof   
                   
Edwin