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After using the Karnaugh map to find the minimal representation of a Boolean function, my answer is 1.

Is 1 a valid answer for minimal representation?

If yes, what is the implication of a Boolean function has 1 as its minimal representation?

WingLeo
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    It is a valid answer, the implication being that your Boolean function outputs $1$ for every input. If you want, edit your post and include the Boolean function so we can check your work. – Ragib Zaman Sep 16 '13 at 08:03
  • Thanks Zaman. here's the BF: ((x'y')(yz))'. Since I'm new here (also in discrete math), i don't know if my notation compiles to the convention of this website or not. My apologies if it is not. – WingLeo Sep 16 '13 at 08:21

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You are correct to find that $((x'y')(yz))'$ has minimal representation $1$ (which means the output is always $1,$ no matter the values of $x,y$ or $z$). To see this, note $(x'y')(yz)$ always outputs $0$, since $x'y'$ requires $y$ to be $0$ to output $1$ and $yz$ requires $y$ to be $1$ to output $1$, and you can't have both of those at the same time. Since $(x'y')(yz)$ always outputs $0,$ $((x'y')(yz))'$ always outputs $1.$

Ragib Zaman
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  • Thanks again, I derive this function from a logic gate. Now my question is: What's the use of such a logic gate if whatever the input is its output is always True? I know this is just an exercise question. But I wonder in real life, what situations this kind of logic gate [the opposite of this one is a logic gate always output False] will be used? I've search online for more than one day and consult my textbook (Epp and Grimaldi). but could not find an answer. wish someone there can help. – WingLeo Sep 16 '13 at 09:54
  • @WingLeo I don't actually know how logic gates are used practically, but I can't imagine this particular example would ever be used. As you said, it's probably just an exercise question with no particular use itself. – Ragib Zaman Sep 16 '13 at 10:46