A property of Boolean algebra

In a Boolean algebra $$\mathcal B$$, we know that $$x+\bar{x}y=x+y\text{ for all } x, y\in \mathcal B.$$ By following the above identity, we can also write $$xy+\bar{x}yz=xy+yz.$$ Can we write $$\bar{y}xz+yp=xz+yp,\text{ where p is distinct from x and z}?$$ Or is there any alternative way to simplify the expression on the left side of the last equation further?

• It would be good if you include in your question the steps to get what you wrote after "By following the above identity". Jun 24, 2021 at 4:25
• @JMP How, it is not true? In that case, both sides of the equation give you $1$.
– gete
Jun 24, 2021 at 5:50
• @JMP These are logical operations and symbols, but not usual addition, multiplication or real number $1$ and $0$. $0$ and $1$ are just symbols.
– gete
Jun 24, 2021 at 6:07
• @JMP You can't remove $z$ from the equation $xyz+x'yz=xyz+yz$ unless the structure is cancelative. However, such an equation will hold according to the second identity.
– gete
Jun 24, 2021 at 9:32
• @JMP His answer is a counter example showing that the last identity doesn't hold. Thereby giving a good hint that the expression $\bar{y}xz+yp$ is in the most simplified form. The second identity which you are mentioning here is an already proven identity.
– gete
Jun 24, 2021 at 9:44

No. Taking $$x = y = z = 1$$ and $$p = 0$$, you get $$\bar yxz +yp = 0$$ but $$xz + yp = 1$$.

• Thank you for the answer. By the ways, do you mean that the expression $\bar{y}xz+yp$ cannot be further simplified?
– gete
Jun 24, 2021 at 4:30
• I don't think that $\bar yxz + yp$ can be simplified. Jun 24, 2021 at 4:37
• Well. That's fine and it got my problem fixed. Thank you.
– gete
Jun 24, 2021 at 4:55