# Proof

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:

s = sage.logic.propcalc.formula("x&(y|z)") s.truthtable()
s = sage.logic.propcalc.formula("(x&y)|(x&z)") s.truthtable()
• The comparison of the last columns of the truth tables shows that $x\wedge (y\vee z)\Longleftrightarrow(x\wedge y)\vee (x\wedge z)$ is a tautology.
• The proof of the tautology $x\vee (y\wedge z)\Longleftrightarrow(x\vee y)\wedge (x\vee z)$ can be done similarily.

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

1. Mendelson Elliott: "Theory and Problems of Boolean Algebra and Switching Circuits", McGraw-Hill Book Company, 1982