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?
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.