SOLUTION: 1. A -> (D -> E) 2. F -> [(A & B) v (A & ~B)] concl: (F & D) -> E Professor says it is 11 lines w/ CP. HELP ME.

Algebra ->  Proofs -> SOLUTION: 1. A -> (D -> E) 2. F -> [(A & B) v (A & ~B)] concl: (F & D) -> E Professor says it is 11 lines w/ CP. HELP ME.       Log On


   

Question 929849: 1. A -> (D -> E)
2. F -> [(A & B) v (A & ~B)]
concl: (F & D) -> E
Professor says it is 11 lines w/ CP.
HELP ME.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
 1. A -> (D -> E)
 2. F -> [(A & B) v (A & ~B)]
 concl: (F & D) -> E

 3.  F -> [A & (Bv~B)]   2,DL     (distributive law, "take out A & "
 4.  F -> A              3,CT     (conjunction with a tautology Bv~B)
 5.  F -> (D -> E)       1,4,HS   (hypothetical syllogism)
 6.  ~F v (D -> E)       5,CD     (conditional to disjunction)
 7.  -F v (~D v E)       6,CD     (conditional to disjunction)
 8.  (~F v ~D) v E       7,       (associative law)
 9.  ~(F & D) v E        8,DM     (deMorgan's law)          
10.  (F & D) -> F        9,CD     (disjunction to conditional)

I did it in 10 lines.
Have you studied "conditional to disjunction" and its
reverse "disjunction to conditional)? Is that what you
label "CP"?  [Some books label differently]. 

Edwin