(related to Lemma: Distributivity of Conjunction and Disjunction)
We want to show that the conjunction "$\wedge$" and the disjunction "$\vee$" are distributive over each other. * It is sufficient to show the following equivalences. $$\begin{array}{rcl} x\wedge (y\vee z)&\Longleftrightarrow&(x\wedge y)\vee (x\wedge z),\\ x\vee (y\wedge z)&\Longleftrightarrow&(x\vee y)\wedge (x\vee z), \end{array}\quad( * )$$ because the equivalence $$\begin{array}{rcl} (y\vee z)\wedge x&\Longleftrightarrow&(y\wedge x)\vee (z\wedge x),\\ (y\wedge z)\vee x&\Longleftrightarrow&(y\vee x)\wedge (z\vee y) \end{array}$$ will follow immediately from the following facts: * conjunction $\wedge$ is associative, * disjunction $\vee$ is associative, * conjunction $\wedge$ is commutative, and * disjunction $\vee$ is commutative. * Using the truth tables of conjunction and disjunction, we construct the truth tables of both sides of the equivalences $( * )$:
| $[[x]]_I$ | $[[y]]_I$ | $[[z]]_I$ | $[[x\wedge (y\vee z)]]_I$ | $[[(x\wedge y)\vee (x\wedge z)]]_I$ |
|---|---|---|---|---|
| $0$ | $0$ | $0$ | $0$ | $0$ |
| $0$ | $0$ | $1$ | $0$ | $0$ |
| $0$ | $1$ | $0$ | $0$ | $0$ |
| $0$ | $1$ | $1$ | $0$ | $0$ |
| $1$ | $0$ | $0$ | $0$ | $0$ |
| $1$ | $0$ | $1$ | $1$ | $1$ |
| $1$ | $1$ | $0$ | $1$ | $1$ |
| $1$ | $1$ | $1$ | $1$ | $1$ |
This can be verified using Sagecell:
