Question 531393
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You'll have to come up with the reasons but it should go something like this:


~(J & K) -> ~J v ~K


~(L & M) -> ~L v ~M


Since J v L, ~(~J & ~L)


~(~J & ~L) -> ~K v ~M -> ~(K & M)


That last step, in English: Since J and L cannot both be false, one or the other or both of K and M must be false, so both K and M cannot be true.  


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