Simplify using boolean algebra laws/formulas

I am trying to learn logic expression simplification using boolean algebra laws/formulas, but I don't understand it at all. We have this expression: $$(x'\wedge y \wedge z' ) \vee (x' \wedge z) \vee (x \wedge y)$$ . In an example solution, we did this:

= $$(x'\wedge y \wedge z' ) \vee (x' \wedge z) \vee (x \wedge y)$$

= $$(x'\wedge y \wedge z' ) \vee (x' \wedge y \wedge z) \vee (x' \wedge y' \wedge z ) \vee (x \wedge y)$$

= $$(x'\wedge y ) \vee (x' \wedge z ) \vee (x \wedge y)$$

= $$y \vee (x' \wedge z )$$

However, I don't understand those steps at all. Could someone explain to me what boolean algebra laws/formulas we did apply at those steps and how? I would be grateful. Thanks!

• The identity $a \equiv (a\wedge b) \vee (a \wedge b')$ is exploited in all lines. On 2nd row, it is used on $(x'\wedge z)$, and the term $x'\wedge y \wedge z$ is also duplicated to use the identity twice on 3rd row. – FormerMath Mar 17 at 17:41

The following general equivalence principles are used:

$$p = (p \land q) \lor (p \land q')$$

$$p = (p \lor q) \land (p \lor q')$$

Idempotence

$$p = p \lor p$$

$$p = p \land p$$

$$(x'\land y \land z' ) \lor (x' \land z) \lor (x \land y) \overset{Adjacency: \ x' \land z = (x' \land y \land z) \lor (x' \land y' \land z)}{=}$$
$$(x'\land y \land z' ) \lor (x' \land y \land z) \lor (x' \land y' \land z) \lor (x \land y) \overset{Idempotence: \ (x' \land y \land z) = (x' \land y \land z) \lor (x' \land y \land z)}{=}$$
$$(x'\land y \land z' ) \lor (x' \land y \land z) \lor (x' \land y \land z) \lor (x' \land y' \land z) \lor (x \land y) \overset{Adjacency: \ (x'\land y \land z' ) \lor (x' \land y \land z) = x' \land y}{=}$$
$$(x'\land y) \lor (x' \land y \land z) \lor (x' \land y' \land z) \lor (x \land y) \overset{Adjacency: \ (x' \land y \land z) \lor (x' \land y' \land z) = x' \land z}{=}$$
$$(x'\land y) \lor (x' \land z) \lor (x \land y) \overset{Adjacency: \ (x'\land y) \lor (x \land y) = y}{=}$$
$$y \lor (x' \land z)$$