Question 1112731
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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
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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