# Simplification of of Boolean expressions

I have been tasked to simplify the following boolean function of three variables $$f(x,y,z) = xy\bar{z}\vee \bar{x}\bar{y}\vee\overline{x\vee y\vee z} \vee xyz$$

The notation is such that $$xy$$ means $$x\wedge y$$. I believe that the most simplified form of the function is $$f(x,y,z) = xyz$$. Here is my unsuccessful attempt at a derivation. I wish to show that the expression excluding $$xyz$$ is $$0$$. \begin{aligned}xy\bar{z}\vee \bar{x}\bar{y}\vee\overline{x\vee y\vee z} &= xy\bar{z}\vee\bar{x}\bar{y}\vee \bar{x}\bar{y}\bar{z}\\ &= xy\bar{z} \vee \bar{x}\bar{y}(1\vee\bar{z})\\&= xy\bar{z} \vee \bar{x}\bar{y}\end{aligned} at this point I am unfortunately stuck. Are there some rules of boolean algebra that I'm breaking?

## 1 Answer

As you continue, \begin{align*} f(x,y,z)&=xy\bar{z}\lor\bar{x}\bar{y}\lor\overline{x\lor y\lor z}\lor xyz \\ &=xy\bar{z}\lor\bar{x}\bar{y}\lor xyz \\ &=xy\bar{z}\lor xyz\lor\bar{x}\bar{y} \\ &=xy(\bar{z}\lor z)\lor\bar{x}\bar{y} \\ &=xy\lor\bar{x}\bar{y}. \end{align*} This cannot be further simplified.

• Thank you. Unfortunately I don't see how you managed to get ride of $\overline{x\vee y\vee z}$. Commented Apr 9, 2021 at 22:28
• @LimChengAng You have already eliminated that term in your post.I just started from where you left. Commented Apr 9, 2021 at 23:06