Question

Prove that the two equations for E in the example starting on page B-7 are equivalent by using DeMorgan’s theorems and the axioms shown on page B-7

Short answer

Prove is provided below by using DeMorgan’s law

Step-by-step solution

Write both equations for E

$$\begin{gathered} E=\left(\left(A.B\right)+\left(A.C\right)+\left(B.C\right)\right).\left(\overline{A.B.C}\right) \\ E=A.B.\overline{C}+A.\overline{B}.C+\overline{A}.B.C \end{gathered}$$

Prove the equation by using DeMorgan’s law

First, we rewrite the last factor

$E=\left(\left(A.B\right)+\left(A.C\right)+\left(B.C\right)\right).\left(\overline{A}+\overline{B}+\overline{C}\right)$

Now using the distributive property

role="math" $E=AB\left(\overline{A}+\overline{B}+\overline{C}\right)+AC\left(\overline{A}+\overline{B}+\overline{C}\right)+BC\left(\overline{A}+\overline{B}+\overline{C}\right)$

Distribute each term

First, we distribute$AB\left(\overline{A}+\overline{B}+\overline{C}\right)$

$$\begin{gathered} =AB\overline{A}+AB\overline{B}+AB\overline{C} \\ =0+0+AB\overline{C} \\ =AB\overline{C} \end{gathered}$$

Now, distribute $AC\left(\overline{A}+\overline{B}+\overline{C}\right)$

$$\begin{gathered} =AC\overline{A}+AC\overline{B}+AC\overline{C} \\ =0+AC\overline{B}+0 \\ =AC\overline{B} \end{gathered}$$

In the last, distribute $BC\left(\overline{A}+\overline{B}+\overline{C}\right)$

$$\begin{gathered} =BC\overline{A}+BC\overline{B}+BC\overline{C} \\ =BC\overline{A}+0+0 \\ =BC\overline{A} \end{gathered}$$

Write the final equation

$E=A.B.\overline{C}+A.\overline{B}.C+\overline{A}.B.C$

Hence proved both the equation are equivalent