Question

Derive the product-of-sums representation for Eshown on page B-11 starting with the sum-of-products representation. You will need to use DeMorgan’s theorems.

Short answer

The product-of -sums representation for E from page B-11 is:

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

The sum of products after applying DeMorgan’s theorems is:

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

Step-by-step solution

Determine the DeMorgan’s Theorem

DeMorgan’s theorem:

• The complement of the product of the terms is equal to the sum of the complement of each term.
• The complement of the sum of the terms is equal to the product of each term.

Determine the product of sums

The product-of -sums representation for E from page B-11 is:

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

Apply DeMorgan’s theorem

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

Apply the DeMorgan’s theorem again,

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

Therefore, the product of sum representation of E is,

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