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I would like to have a step-by-step simplification of this boolean expression $\bar{x}\bar{y}\bar{z}\bar{w}+\bar{x}\bar{y}z\bar{w}+\bar{x}yz\bar{w}+\bar{x}yzw+x\bar{y}\bar{z}\bar{w}+x\bar{y}z\bar{w}+x\bar{y}zw+xy\bar{z}w+xyzw? $

I have done:

$\bar{x}\bar{y}\bar{w}(\bar{z}+z)+\bar{x}yz(\bar{w}+w)+x\bar{y}\bar{z}\bar{w}+x\bar{y}z(\bar{w}+w)+xyz(\bar{w}+w)=$ $=\bar{x}\bar{y}\bar{w}+\bar{x}yz+x\bar{y}\bar{z}\bar{w}+x\bar{y}z+xyz=$

$=xyz+x\bar{y}z+\bar{x}yz+\bar{y}\bar{w}(\bar{x}+x\bar{z})$

But I don't know how to reduce furthermore

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  • $\begingroup$ What is $x$ and $\overline{x}$? $\endgroup$ – Dietrich Burde Apr 17 at 15:21
  • $\begingroup$ is like NOT x, or x! $\endgroup$ – Mathematics1990 Apr 17 at 15:23
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    $\begingroup$ What have you done yourself sofar? $\endgroup$ – nilo de roock Apr 17 at 15:24
  • $\begingroup$ Let me get you started $\bar{x}\bar{y}\bar{z}\bar{w}+\bar{x}\bar{y}z\bar{w}=\bar{x}\bar{y}\bar{w}(\bar{z}+z)=\bar{x}\bar{y}\bar{w}$. $\endgroup$ – kingW3 Apr 17 at 15:30
  • $\begingroup$ Two basic approaches here Karnaugh Map or De Morgan. Lets see your effort first. $\endgroup$ – Warren Hill Apr 17 at 15:37

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