Question

1. Show that \({\bf{(\bar 1}} \cdot {\bf{\bar 0) + (1}} \cdot {\bf{\bar 0) = 1}}\).
2. Translate the equation in part (a) into a propositional equivalence by changing each \(0\) into an \({\bf{F}}\), each \(1\) into a \({\bf{T}}\), each Boolean sum into a disjunction, each Boolean product into a conjunction, each complementation into a negation, and the equals sign into a propositional equivalence sign.

Short answer

a) The given \((\bar 1 \cdot \bar 0) + (1 \cdot \bar 0) = 1\) is proved.

b) The translated equation of the given is \(({\bf{\neg T}} \wedge {\bf{\neg F}}) \vee ({\bf{T}} \wedge {\bf{\neg F}}) \equiv {\bf{T}}\).

Step-by-step solution

Definition

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

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.

(a) Using the Boolean product and sum, definition of complement

By the definition of complement, Boolean product and sum

\(\begin{aligned} (\bar 1 \cdot \bar 0) + (1 \cdot \bar 0) &= (0 \cdot \bar 0) + (1 \cdot \bar 0)\\ & = (0 \cdot 1) + (1 \cdot \bar 0)\\ & = (0 \cdot 1) + (1 \cdot 1)\\ & = 0 + (1 \cdot 1)\\ &= 0 + 1\\ &= 1\end{aligned}\)

Therefore, you get \((\bar 1 \cdot \bar 0) + (1 \cdot \bar 0) = 1\).

(b) Using the Boolean product and sum

\((\bar 1 \cdot \bar 0) + (1 \cdot \bar 0) = 1\)

Replace 0 by F

Replace 1 by T

Replace + by \( \vee \)

Replace \( \cdot \) by \( \wedge \)

Replace - by \(\neg \) (placed in front of the expression that used to be underneath)

Replace = by \( \equiv \)

Therefore, you get \(({\bf{\neg T}} \wedge {\bf{\neg F}}) \vee ({\bf{T}} \wedge {\bf{\neg F}}) \equiv {\bf{T}}\).