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This is for a class on logic gates and boolean expressions, the expression I was given is the following:

$$x'y'z'+w'x'yz'+wx'yz'$$

I have done one other question like this so far and I struggled with it, so far I have only been able to do the following:

$$x'z'(y'+w'y+wy)$$

I'm not entirely sure if I'm even allowed to remove the common elements the way I did here but it was the first thing I noticed, after that I did the following:

$$x'z'(y' + y)$$

I know that $w'y+wy = y$ so I placed that there and I know that $y + y'$ is always true so the final answer I arrived at was:

$$ x'z'$$

I don't know exactly if everything I did is allowed, and I couldn't find an answer online either. Could someone confirm if I was able to do this question properly or point out to me where I messed up. Thank you very much

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    $\begingroup$ Everything seems right. $\endgroup$ – Hugo Oct 8 '18 at 23:00
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    $\begingroup$ Your reduction is fine. To make things easier on yourself, instead of trying to factor the whole thing at once, first try looking for pairs of terms with common subterms that are each other's negation. In this case, you have $w'x'yz'$ and $wx'yz'$, which you can then combine to eliminate $w$. $\endgroup$ – amd Oct 8 '18 at 23:11
  • $\begingroup$ @amd if I were to combine w'x'yz' and wx'yz' to eliminate what would the final product be? would it just be = x'y'z' + x'z' + x'z' ? $\endgroup$ – AFC Oct 9 '18 at 0:31
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    $\begingroup$ Combining the two terms gives you $(w+w')x'yz'=x'yz'$, so the original formula becomes $x'y'z'+x'yz'$. Continue the process from there. $\endgroup$ – amd Oct 9 '18 at 0:33
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You correctly notice you can collect $x'z'$; this leaves to simplify $$ y'+w'y+wy=y'+y(w'+w)=y'+y1=y'+y=1 $$ so what remains is indeed $x'z'$.

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