Question 251008
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Note that [(A & B) v (A & ~B)] is equivalent to A.  So


F -> [(A & B) v (A & ~B)] is the same thing as F -> A, hence A -> (D -> E) is the same thing as F -> (D -> E), and (F & D) -> E follows directly.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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