I am conflicted because our professor said that "(not) A implies B only and only if A and B = 0" which doesn't match with what I found on Wikipedia or other books on Boolean algebra: "A implies B = (not) A or B". Are the two equivalent, or one of them is wrong? Our professor also said that the imply operator is similar to a negated AND...
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Have you been introduced to De Morgan's laws? – Schala Oct 10 '15 at 18:05
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Also, have you written out truth tables for each statement? At the least, you can try out some values: $A = 0$ and $B = 1$ – Schala Oct 10 '15 at 18:07
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the tables are different that is the reason I am asking this question... – Andrew Oct 10 '15 at 18:11
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Then it seems to me that you already have the answer to this question. – Schala Oct 10 '15 at 18:13
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(not A) => B if only false when A is true and B is false. So if A and B are both true (not A) => B is true. But A and B = 0 is false. So these are not equivalent.
not (A => B) is only true when (A => B) is false. (A => B) is only false when A is true and B is false. So if A and B are both false, A and B = 0 is true but not(A => B) is false.
You must be misquoting your professor.
A or B = 0 is only true if A and B are both false. (not) A => B is only true if A and B are both false.
So that must be what your professor meant.
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