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How do I simplify the following expression with Boolean Algebra? Please show what you used to simplify so I can understand.

$$ABC + AB'C' + ABC' + A'B'C'$$

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    $\begingroup$ I would go with $ABC + AB'C' + ABC' + A'B'C' = AB + B'C'$, since $A+A'=1$. $\endgroup$ – Mario Carneiro Jun 4 '13 at 4:49
  • $\begingroup$ $$ABC+ABC'+AB'C'+A'B'C'=AB(C+C')+B'C'(A+A')=AB+B'C'$$ $\endgroup$ – lab bhattacharjee Jun 4 '13 at 4:50
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First I want to group the elements that are similar. This will allow me to start reducing the expression.

$$ABC + AB'C' + ABC' + A'B'C'$$ $$ABC + ABC' + AB'C' + A'B'C'$$ $$[AB(C+C')] + [B'C'(A+A')] // Group.$$ $$AB + B'C' // α+α' = 1; α * 1 = α $$

Edit: For me Boolean Algebra is superior, but if you ever get stuck on a problem try using a Karnaugh map: http://en.wikipedia.org/wiki/Karnaugh_map

If you read the Wiki page you will see that with a Karnaugh map you can simplify this expression.

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  • $\begingroup$ Liking this Karnaugh_map. $\endgroup$ – katrina Jun 4 '13 at 5:08
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Note that you have both $ABC$ and $ABC'$:

$$ABC+ABC'=AB(C+C')=AB\;.$$

The rest is $AB'C'+A'B'C'$, and you can use the same idea:

$$AB'C'+A'B'C'=(A+A')B'C'=B'C'\;.$$

After those simplifications the expression has been reduced to $AB+B'C'$, which is as simple as it’s going to get.

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Remember that $C+C'=1$ and $X\cdot1=X$, and $X(Y+Z)=XY+XZ$. Then we have:

$$\begin{align} &ABC+AB'C'+ABC'+A'B'C'=\\ &ABC+ABC'+AB'C'+A'B'C'=\\ &AB(C+C')+(A+A')B'C'=\\ &&=AB+B'C' \end{align}$$

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