Question

Show that you obtain the absorption laws for propositions (in Table \({\bf{6}}\) in Section \({\bf{1}}{\bf{.3}}\)) when you transform the absorption laws for Boolean algebra in Table 6 into logical equivalences.

Short answer

The transformed form of the absorption law for Boolean algebra into logical equivalences is \(x \vee \left( {x \wedge y} \right) \equiv x,x \wedge \left( {x \vee y} \right) \equiv x\).

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 or \(AND\) is \(1\) if both terms are \(1\) .

Absorption laws (Boolean algebra)

\(\begin{array}{c}x + xy = x\\x\left( {x + y} \right) = x\end{array}\)

Absorption laws (Logical equivalences)

\(\begin{array}{l}p \vee \left( {p \wedge q} \right) \equiv p\\p \wedge \left( {p \vee q} \right) \equiv p\end{array}\)

Using the first absorption law

\(x + xy = x\)

Note: \(xy\) implies \(x \times y\).

\(x + \left( {x \times y} \right) = x\)

Transform to logical equivalence

Replace \( \times \) by \( \wedge \)

Replace \( + \) by \( \vee \)

Replace \( = \) by \( \equiv \)

\({\bf{x}} \vee {\bf{(x}} \wedge {\bf{y)}} \equiv {\bf{x}}\)

Therefore, it gets the first absorption law for logical equivalences.

Using the second absorption law

\(x\left( {x + y} \right) = x\)

Note: \(x\left( {x + y} \right)\) implies \(\left( {x \times \left( {x + y} \right)} \right)\).

\(x \times \left( {x + y} \right) = x\)

Transform to logical equivalence

Replace \( \times \) by \( \wedge \)

Replace \( + \) by \( \vee \)

Replace \( = \) by \( \equiv \)

\({\bf{x}} \wedge ({\bf{x}} \vee {\bf{y}}) \equiv {\bf{x}}\)

Therefore, it gets the second absorption law for logical equivalences.