SOLUTION: A -> (B & C) C -> (E -> I) ~(D -> I) / therefore, (A & G) -> (~I & ~E)

Question 971097: A -> (B & C)
C -> (E -> I)
~(D -> I) / therefore, (A & G) -> (~I & ~E)
Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!

 1.  A -> (B & C)
 2.  C -> (E -> I)
 3.  ~(D -> I)	
        / therefore, (A & G) -> (~I & ~E)

 4. (A -> B) & (A -> C)          1, composition
 5. A -> C                       4, simplification
 6. A -> (E -> I)              2,5, hypothetical syllogism
 7. (A & E) -> I                 6, exportation
 8. ~I -> ~(A & E)               7, contrapositive
 9. ~(~D v I)                    3, material implication
10. ~~D & ~I                     9, DeMorgan's law
11. D & ~I                      10, double negation
12. ~I                          11, simplification
13. ~(A & E)                  8,12, hypothetical syllogism       
14. ~A v ~E                     13, DeMorgan's law
15. A -> ~E                     14, material implication
16. (A -> ~E) v ~G              15, addition
17. (~A v ~E) v ~G              16, material implication 
18. ~A v (~E v ~G)              17, association
19. ~A v (~G v ~E)              18, commutation
20. (~A v ~G) v ~E              19, association
21. ~(A & G) v ~E               20, DeMorgan's law
22. (A & G) -> ~E               21, exportation
23. ~E -> ~E & ~I               12, natural deduction
24. (A & G) -> (~E & ~I)     22,23, hypothetical syllogism

Edwin

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