Question

In Exercises 35–42, use the laws in Definition \(1\) to show that the stated properties hold in every Boolean algebra.

35. Show that in a Boolean algebra, the idempotent laws \(x \vee x{\bf{ = }}x\) and \(x \wedge x{\bf{ = }}x\) hold for every element \(x\).

Short answer

By using the identity, complement and distributive law, you get \(x \vee x = x,x \wedge x = x\).

Step-by-step solution

Definition

The complement of an element: \(\bar 0 = 1\) and \(\bar 1 = 0\)

The Boolean sum \( + \) or \(OR\) is \(1\) if either term is \(1\).

The Boolean product \( \cdot \) or \(AND\) is \(1\) if both terms are \(1\).

Identity laws

\(\begin{array}{c}x \vee 0 = x\\x \wedge 1 = x\end{array}\)

Complement laws

\(\begin{array}{c}x \vee \bar x = 1\\x \wedge \bar x = 0\end{array}\)

Distributive laws

\(\begin{array}{c}x \vee \left( {y \wedge z} \right) = \left( {x \vee y} \right) \wedge \left( {x \vee z} \right)\\x \wedge \left( {y \vee z} \right) = \left( {x \wedge y} \right) \vee \left( {x \wedge z} \right)\end{array}\)

Idempotent laws

\(\begin{array}{c}x \vee x = x\\x \wedge x = x\end{array}\)

Using the identity, complement and distributive laws

To proof: \(x \vee x = x\)

Using the identity, complement and distributive laws

\(\begin{eqnarray}x \vee x &=& \left( {x \vee x} \right) \wedge 1\\ &=& \left( {x \vee x} \right) \wedge \left( {x \vee \bar x} \right)\\ &=& x \vee \left( {x \wedge \bar x} \right)\\ &=& x \vee 0\\ &=& x\end{eqnarray}\)

Therefore, it gets \(x \vee x{\bf{ = }}x\).

Using the identity, complement and distributive laws

To proof: \(x \wedge x = x\)

Using the identity, complement and distributive laws

\(\begin{eqnarray}x \wedge x &=& \left( {x \wedge x} \right) \vee 0\\ &=& \left( {x \wedge x} \right) \vee \left( {x \wedge \bar x} \right)\\ &=& x \wedge \left( {x \vee \bar x} \right)\\ &=& x \wedge 1\\ &=& x\end{eqnarray}\)

Therefore, it gets \(x \wedge x{\bf{ = }}x\).