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!

  • $\begingroup$ 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. $\endgroup$
    – FormerMath
    Commented Mar 17, 2019 at 17:41

1 Answer 1


The following general equivalence principles are used:


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

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


$p = p \lor p$

$p = p \land p$

Applied to your statement:

$ (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}{=}$

$y \lor (x' \land z)$


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