Simplify the following expression: (A + B + 1) . (A' +B' + 1) . (A + B + C) .…

Step 1: Simplify terms involving the constant '1'

In Boolean algebra, the Identity Law states that X + 1 = 1.

• The first term (A + B + 1) simplifies to 1.

• The second term (A' + B' + 1) simplifies to 1.

The expression now becomes: 1 \cdot 1 \cdot (A + B + C) \cdot (A' + B' + C')

Since 1 \cdot X = X, we are left with:

(A + B + C) \cdot (A' + B' + C')

Step 2: Expand the remaining terms

We distribute (A + B + C) over (A' + B' + C'): (A \cdot A') + (A \cdot B') + (A \cdot C') + (B \cdot A') + (B \cdot B') + (B \cdot C') + (C \cdot A') + (C \cdot B') + (C \cdot C')

Step 3: Apply the Complement Law

The Complement Law states that X \cdot X' = 0.

• A \cdot A' = 0

• B \cdot B' = 0

• C \cdot C' = 0

The expression simplifies to:

AB' + AC' + BA' + BC' + CA' + CB'

Step 4: Group and Factor

Group the terms to match the requested format:

(AB' + CB') + (BA' + CA') + (AC' + BC')

Factor out the common variables:

B'(A + C) + A'(B + C) + C'(A + B)

Final Answer

The simplified expression is [(A + C)B'] + [(B + C)A'] + [(A + B)C'], which corresponds to Option 2.