Boolean algebra: Why does the equation hold? I want to show that $$\bar y\land (x\lor z)\land (\bar x \lor  \bar y) = \bar y\land (x\lor z)$$ 
I have done the following:  \begin{align*}\bar y\land (x\lor z)\land (\bar x \lor \bar y) &=\bar y\land \left [(x\lor z)\land (\bar x \lor \bar y)\right ]\\ & =\bar y\land \left [\left ((x\lor z)\land \bar x\right )\lor \left ((x\lor z)\land \bar y\right )\right ] \\ & = \bar y\land \left [\left (\left (x\land \bar x\right )\lor \left (z\land \bar x\right )\right )\lor \left (\left ( x\land \bar y\right )\lor \left (z\land \bar y\right )\right )\right ] \\ & = \bar y\land \left [\left (0\lor \left (z\land \bar x\right )\right )\lor \left (\left ( x\land \bar y\right )\lor \left (z\land \bar y\right )\right )\right ] \\ & = \bar y\land \left [\left (z\land \bar x\right )\lor \left ( x\land \bar y\right )\lor \left (z\land \bar y\right )\right ]\end{align*} Is everything correct so far? How can we continue? 
 A: You want to show $$\bar y\land (x\lor z)\land (\bar x \lor  \bar y) = \bar y\land (x\lor z)$$
Intuitively, the left hand side of this says that $y$ must be false (because we have $\bar y \land \dots$), and so $(\bar x \lor \bar y)$ is a tautology—it is always true. So we can rewrite this as $\bar y \land (x \lor z)$, which is exactly the right hand side. So they are equivalent statements.
More rigorously,
$$
\begin{align}
\bar y \land (x \lor z) \land (\bar x \lor \bar y) \\
=\bar y \land (\bar x \lor \bar y) \land (x \lor z) && \text{commutativity} \\
=\bar y \land (x \lor z) && \text{absorption} \\
\end{align}
$$
The ‘absorption’ law we just used is a standard identity but if you really want proof,
$$
\begin{align}
x \land (x \lor y) \\
=(x \land x) \lor (x \land y) && \text{distributive} \\
=x \lor (x \land y) && \text{idempotent} \\
=(x \land 1) \lor (x \land y) && \text{identity} \\
=x \land (1 \lor y) && \text{distributive} \\
=x \land 1 && \text{annulment} \\
=x && \text{identity} \\
\end{align}
$$
