I started with (AB' + A'B)' and ended up with (A'B' + AB). Is this all the farther I can go? I feel like this is always going to be true, but I'm not sure how to prove it algebraically.
Can (x'y' + xy) be simplified?
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$\begingroup$
boolean-algebra
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0$(fg)'=fg'+f'g, (fg)''=f''+2f'g+2fg'+g''$ etc. what are $x$ and $y$? what are you asking? – 2011-04-13
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0sorry, should have used A and B instead of x and y...A and B are booleans. – 2011-04-13
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1What do you mean by simpler? Fewer gates (AND/NOT/OR)? Smaller length when viewed as a string? – 2011-04-13
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2In a sense, this is the farthest you can go. What you have reached is the equivalence operation, which is the negation of XOR. – 2011-04-13
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1@Brandon: You should probably write that as an answer. – 2011-04-13
1 Answers
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As per request, I am making my comment into an answer.
In a sense, this is the farthest you can go. What you have reached is the equivalence operation, which is the negation of XOR. Equivalence, sometimes denoted XNOR, returns true if the inputs are either both true or both false. See the Wikipedia page for XNOR for more information.
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1I have always thought NXOR would be a more logical name than XNOR – 2011-04-13
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0thanks, now i see this is an xnor. – 2011-04-13
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0@Henry: NXOR would be more logical, but does not have an easy pronunciation like XNOR does. – 2011-04-17
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0"Nex or"? `:)` – 2011-04-26
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0@Henry I've always thought "=" should be used. – 2015-01-16