Simplify Boolean Expression

The task is to simplify the following Boolean expression:

$$ (AB + C + DC)(AC + BC + D) $$

Boolean Expression Simplification Steps

Simplify First Parenthesis

Let's analyze the first part of the expression: $ (AB + C + DC) $. Using the commutative law, we know that $ DC $ is equivalent to $ CD $. So the expression becomes $ (AB + C + CD) $. Now, consider the terms $ C + CD $. We can use the distributive law by factoring out $ C $: $ C + CD = C(1 + D) $. According to the identity law in Boolean algebra, $ 1 + D = 1 $. Substituting this back, we get $ C(1) = C $. Thus, the first parenthesis $ (AB + C + CD) $ simplifies to $ (AB + C) $.

Expression Expansion

The expression is now reduced to:

$$ (AB + C)(AC + BC + D) $$

We apply the distributive law $ X(Y + Z) = XY + XZ $ to expand this:

$$ AB(AC + BC + D) + C(AC + BC + D) $$

Simplify Expanded Terms

Now, let's simplify each part of the expanded expression:

Part 1: Expanding $ AB(AC + BC + D) $

Using the distributive law:

$$ AB \cdot AC + AB \cdot BC + AB \cdot D $$

Applying the associative law and the idempotent law ($ X \cdot X = X $):

$$ A \cdot B \cdot A \cdot C = A \cdot A \cdot B \cdot C = A \cdot B \cdot C = ABC $$

$$ A \cdot B \cdot B \cdot C = A \cdot B \cdot C = ABC $$

$$ A \cdot B \cdot D = ABD $$

So, $ AB(AC + BC + D) $ simplifies to $ ABC + ABC + ABD $. Using the idempotent law ($ X + X = X $), this becomes $ ABC + ABD $.

Part 2: Expanding $ C(AC + BC + D) $

Using the distributive law:

$$ C \cdot AC + C \cdot BC + C \cdot D $$

Applying the idempotent law ($ C \cdot C = C $):

$$ C \cdot A \cdot C = A \cdot C \cdot C = A \cdot C = AC $$

$$ C \cdot B \cdot C = B \cdot C \cdot C = B \cdot C = BC $$

$$ C \cdot D = CD $$

So, $ C(AC + BC + D) $ simplifies to $ AC + BC + CD $.

Combine Simplified Terms

Now, we combine the simplified results from both parts:

$$ E = (ABC + ABD) + (AC + BC + CD) $$

$$ E = ABC + ABD + AC + BC + CD $$

Final Boolean Simplification

We can apply the absorption law, which states $ X + XY = X $.

Consider the term $ ABC $. It can be absorbed by $ AC $ because $ AC + ABC = AC(1 + B) = AC(1) = AC $.

Similarly, $ ABC $ can be absorbed by $ BC $ because $ BC + ABC = BC(1 + A) = BC(1) = BC $.

Since both $ AC $ and $ BC $ are present in the expression $ E = ABC + ABD + AC + BC + CD $, the term $ ABC $ is redundant.

Therefore, the expression simplifies to:

$$ E = AC + BC + ABD + CD $$

Since $ DC $ is the same as $ CD $, the final simplified expression is:

$$ AC + BC + DC + ABD $$