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:

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

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

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

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:

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

Apply DeMorgan’s theorem

Apply the DeMorgan’s theorem again,

role="math" localid="1655292725695" $\begin{array}{l}\mathrm{E}=\overline{(\overline{\mathrm{A}}+\overline{\mathrm{B}}+\mathrm{C})}+\overline{(\overline{\mathrm{A}}+\overline{\mathrm{C}}+\mathrm{B})}+\overline{(\overline{\mathrm{B}}+\mathrm{C}+\mathrm{A})}\\ \mathrm{E}=(\mathrm{A}.\mathrm{B}.\overline{\mathrm{C}})+(\mathrm{A}.\mathrm{C}.\overline{\mathrm{B}})+(\mathrm{B}.\stackrel{}{\overline{\mathrm{C}}.}\overline{\mathrm{A}})\end{array}$

Therefore, the product of sum representation of E is,

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