SOLUTION: 1) (A&B) > ~(S v T) 2) ~E 3) (S v T) v ~(~E & ~F) 4) (~E v F) > (A&B) Conclusion: E v F

Algebra ->  Proofs -> SOLUTION: 1) (A&B) > ~(S v T) 2) ~E 3) (S v T) v ~(~E & ~F) 4) (~E v F) > (A&B) Conclusion: E v F      Log On


   



Question 1112731: 1) (A&B) > ~(S v T)
2) ~E
3) (S v T) v ~(~E & ~F)
4) (~E v F) > (A&B)
Conclusion: E v F

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!

Since ~E is given, the only way for the conclusion to be true is if F is true. Assume ~F. Assuming ~F and given ~E, (~E & ~F) must be true, hence ~(~E & ~F) is false. That leads to (S v T) true by Disjunctive Syllogism applied to (S v T) v ~(~E v ~F). But (S v T) true means ~(S v T) false, hence (A & B) is false by Modus Tollens applied to (A & B) -> ~(S v T). But (A & B) false means (~E v F) is false by Modus Tollens applied to (~E v F) -> (A & B). (~E v F) false means ~(~E v F) is true, which is the same as E & ~F by De Morgan. From E & ~F we get E by Conjunction Elimination (aka Simplification). But we were given ~E, so assuming ~F leads to the contradiction ~E & E. Hence ~F is false, hence F. Finally E v F by Disjunction Introduction (aka Addition)

John

My calculator said it, I believe it, that settles it