Can Known Conditions be Used to Make Assumptions About Equality in Boolean Algebra?

Question

I'm attempting to prove equality between

$(x+z)\bullet(\neg x + y) \bullet (y+z) = y \bullet z$

With the known conditions $x \bullet y=0$ and $x+y=1$.

Using the known conditions, we know that one variable, $x$ or $y$, must be equal to $1$ and the other equal to $0$. We do not know which one is equal to what, just that their values are independent and opposite.

If this is true, does it hold that $\neg x = y$ and $\neg y = x$?

If say, $x = 1$ and $y = 0$, the above conditions are satisfied. Then $\neg x = \neg 1 = 0 = y$ and $\neg y = \neg 0 = 1 = x$, right? Therefore, $y \bullet z = \neg x \bullet z$?

If this is a valid logical deduction, does it have a name?

Notation:

$+$ is equal to OR

$\bullet$ is equal to AND

$\neg$ is equal to NOT

Answer 1

You seem to be misinterpreting the known conditions. We cannot conclude that one of them is $0$ and the other is $1$. Rather, we can conclude that $y=\neg x,$ as the known conditions are the defining properties of $y=\neg x$.

Answer 2

\begin{eqnarray*} \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline x & y & z & & x+z & \neg x+y & y+z & \bullet & & y \bullet z \\ \hline 0&1&0& &0 &1&1&\color{blue}{0}& &\color{blue}{0} \\ \hline 1&0&0& &1 &0&0&\color{blue}{0}& &\color{blue}{0} \\ \hline 0&1&1& &1 &1&1&\color{blue}{1}& & \color{blue}{1} \\ \hline 1&0&1& &1 &0&1&\color{blue}{0}& & \color{blue}{0} \\ \hline \end{array} \end{eqnarray*}