# How can this boolean expression be simplified?

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

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