Finding the Canonical Product of Sums Form for a Boolean Function

This solution explains how to determine the Canonical Product of Sums (POS) form for the given Boolean function (X + Y)(X + Z')(Y + Z).

Understanding Canonical Product of Sums (POS) Form

A Boolean function is in Canonical POS form if it meets these criteria:

• The function is expressed as a product (AND operation) of several Sum terms.
• Each Sum term must contain exactly one instance of every variable involved in the function, either in its normal or complemented form. For a function with $n$ variables, each Sum term must have length $n$.

The specific Boolean function we are analyzing, (X + Y)(X + Z')(Y + Z), involves three variables: X, Y, and Z. Therefore, the Canonical POS form requires each sum term to contain X, Y, and Z.

Converting the Given POS Expression to Canonical Form

The initial expression is:

(X + Y)(X + Z')(Y + Z)

To convert this into the Canonical POS form, we need to modify each sum term so it includes all three variables (X, Y, Z). We utilize the Boolean algebra identity $A = A + 0$. Since $B \cdot B' = 0$ for any variable B, we can add $B \cdot B'$ to a term without changing its value. We then use the distributive law $A + (B \cdot C) = (A + B)(A + C)$ to expand the term.

Step 1: Convert the term $(X + Y)$

This term is missing the variable Z. We introduce $(Z \cdot Z')$:

$$ (X + Y) = (X + Y + Z \cdot Z') $$

Now, apply the distributive law to expand:

$$ (X + Y + Z \cdot Z') = (X + Y + Z)(X + Y + Z') $$

These two sum terms, $(X + Y + Z)$ and $(X + Y + Z')$, both contain all variables and are equivalent to the original $(X + Y)$ term.

Step 2: Convert the term $(X + Z')$

This term is missing the variable Y. We introduce $(Y \cdot Y')$:

$$ (X + Z') = (X + Y \cdot Y' + Z') $$

Apply the distributive law to expand:

$$ (X + Y \cdot Y' + Z') = (X + Y + Z')(X + Y' + Z') $$

These two sum terms, $(X + Y + Z')$ and $(X + Y' + Z')$, are equivalent to the original $(X + Z')$ term and include all variables.

Step 3: Convert the term $(Y + Z)$

This term is missing the variable X. We introduce $(X \cdot X')$:

$$ (Y + Z) = (X \cdot X' + Y + Z) $$

Apply the distributive law to expand:

$$ (X \cdot X' + Y + Z) = (X + Y + Z)(X' + Y + Z) $$

These two sum terms, $(X + Y + Z)$ and $(X' + Y + Z)$, are equivalent to the original $(Y + Z)$ term and include all variables.

Step 4: Combine the converted terms

The original function $F$ is the product of the original terms. We replace each original term with its expanded canonical form:

$$ F = [(X + Y + Z)(X + Y + Z')] \cdot [(X + Y + Z')(X + Y' + Z')] \cdot [(X + Y + Z)(X' + Y + Z)] $$

Step 5: Simplify the resulting expression

We simplify the combined expression using the Boolean algebra property $A \cdot A = A$ (idempotent law). This law states that ANDing a term with itself results in the same term.

Observe the terms in the expanded expression:

• $(X + Y + Z)$ appears twice.
• $(X + Y + Z')$ appears twice.

Applying the idempotent law, we eliminate the duplicate terms:

$$ F = (X + Y + Z)(X + Y + Z')(X + Y' + Z')(X' + Y + Z) $$

This final expression consists of four sum terms. Each term contains all three variables (X, Y, Z), and the function is a product of these terms. Thus, it is the Canonical Product of Sums form.

Final Canonical POS Form Identification

The Canonical Product of Sums form of the given Boolean function (X + Y)(X + Z')(Y + Z) is:

$$ (X + Y + Z)(X + Y + Z')(X + Y' + Z')(X' + Y + Z) $$

This result matches the expression found in Option 1.