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I have this question that I need some help with, I just can't get to grips with simplifying. I'm looking at the rules and such but I just can't see where to apply them. Can someone show me the simplification steps to this so I can answer the rest of my question booklet?

!A.!B.C + A.!B.!C + A.!B.C + A.B.C

Thanks

Karlo
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MrTurvey
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3 Answers3

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There are only three symbols here, so we can use a Karnaugh map.

$\def\X{\times}$ $$ \begin{array}{c|cccc} & AB & A\bar B & \overline{AB} & \bar AB \\ \hline C & \X & \X & \X & \\ \bar C & & \X && \end{array}$$

Here I've put $\X$ in each box for which the formula is true. The Karnaugh map method says that if we have a line of four $\X$es, or a square of four $\X$es, we can write those four $\X$es very simply. We don't have that, but we can do the same for lines of two $\X$es.

The four $\X$es together can be covered by three pairs of two $\X$es. For example, the first two $\X$es in the top row are a pair, and can be represented by the formula $AC$. The second and third $\X$es in the top row are also a pair, covered by the formula $\bar BC$. And the pair of $\X$es in the second column are covered by the formula $A\bar B$. So one way to simplify the formula is $$AC\lor \bar BC\lor A\bar B$$ or in your notation, 

A.C + !B.C + A.!B .

We could also view the three $\X$es in the top row as a row of four that is missing one. The row of four is simply $C$. The row of four, plus the two $\X$es in the second column, is $C\lor A\bar B$. This formula includes an $\X$ in the upper-right box that shouldn't be there, so we should remove it, and get $$(C\lor A\bar B)\land \lnot\left(\bar AB\right).$$ In your notation this is 

(C + A.!B) . !(!A.B)  .

Whether you consider this simpler than the other one depends on what you mean by “simpler”.

MJD
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We have

!A.!B.C + A.!B.!C + A.!B.C + A.B.C

First, we translate it to 'normal' symbols (i.e., the ones I'm used to): $$ (\lnot A\land\lnot B\land C)\lor(A\land\lnot B\land\lnot C)\lor( A\land\lnot B\land C)\lor(A\land B\land C) $$ The first and third part with $\lnot B\land C$ can be taken together, and the $A$ in the two remaining parts can be extrated: $$ (\lnot A\land\lnot B\land C)\lor(A\land\lnot B\land\lnot C)\lor( A\land\lnot B\land C)\lor(A\land B\land C)\\ ((\lnot A \lor A)\land\lnot B\land C)\lor(A\land\lnot B\land\lnot C)\lor(A\land B\land C)=\\ (\lnot B\land C)\lor(A\land\lnot B\land\lnot C)\lor(A\land B\land C)=\\ (\lnot B\land C)\lor(A\land((\lnot B\land\lnot C)\lor( B\land C))=\\ (\lnot B\land C)\lor(A\land(B\implies C)) $$ The replacement by the 'imply' can be made, because the case $\lnot B \land C$ is already true.

There are usually multiple solutions to this kind of problem I think, and the one I found myself is different from this one. (I found this one using Mathematica.)

Ragnar
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$!A.!B.C + A.!B.!C + A.!B.C + A.B.C $

$A*!B*(C+!C) + C(!A.!B+ A.B) $ distributive law

$A*!B*1 + C(!(A+B)+ A.B) $ Unit property and De Morgan’s laws

$A*!B+C(!(A+B)+ A.B)$ Identity laws

JulianP
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