Question

Simplify the following Boolean expression: \((a+b)(c+a+\bar{h})(\bar{h}+c)(a c h+\bar{a} \bar{h})+a h c d+\bar{c}+\bar{h}\)

Short answer

The simplified Boolean expression is: \((ac+\bar{h}a)+\bar{h}b\bar{a}+a h c d+\bar{c}+\bar{h}\).

Step-by-step solution

Apply Distributive Law to (a+b)

We will first apply the Distributive Law to the \((a+b)\) term, resulting in the following expression: \[ac(a+b)+\bar{h}(a+b)(a c h+\bar{a} \bar{h})+a h c d+\bar{c}+\bar{h}\]

Distribute (a+b) further

Distribute \((a+b)\) in the previous expression: \[aca+acb+\bar{h}a(a+b)(a c h+\bar{a} \bar{h})+\bar{h}b(a c h+\bar{a} \bar{h})+a h c d+\bar{c}+\bar{h}\]

Simplify each term with rules

Now, we can apply Boolean algebra rules to simplify each term in the expression: 1. \(aca = ac\), because \(xx = x\). 2. Since \(acb = ab*c =-ac\), we can apply 3. Identity Law (\(x+0 = x\)) to obtain \(ac+0 = ac\). 3. Distribute \(\bar{h}a(a+b)\) as \(\bar{h}a\)(a + b), which becomes \(\bar{h}a^2+\bar{h}ab = \bar{h}a+0\). 4. Distribute \(\bar{h}b(a c h+\bar{a} \bar{h})\) as \(\bar{h}b(a c h+\bar{a} \bar{h}) = \bar{h}bac*h+\bar{h}b\bar{a}\bar{h}\) and simplify it as \(\bar{h}acb+\bar{h}b\bar{a}\bar{h} = 0+ \bar{h}b\bar{a}\bar{h} = \bar{h}b\bar{a}\), since \(0+x=x\). The resulting expression is: \(ac+\bar{h}a+\bar{h}b\bar{a}+a h c d+\bar{c}+\bar{h}\)

Factor out common terms

We notice that (ac\) and \(\bar{h}a\) are common terms. Factor them out, and we obtain: \((ac+\bar{h}a)+\bar{h}b\bar{a}+a h c d+\bar{c}+\bar{h}\)

Simplify and rewrite the expression

The final simplified expression is: \((ac+\bar{h}a)+\bar{h}b\bar{a}+a h c d+\bar{c}+\bar{h}\).

Key concepts

Distributive Law

Understanding the Distributive Law is key to simplifying complex Boolean expressions. In Boolean algebra, the Distributive Law allows us to simplify expressions by spreading out the terms. This law states that for all Boolean variables, the equation \( A(B + C) = AB + AC \) holds. In simpler terms, it means that when an AND operation \(A\) is distributed over an OR operation \(B + C\), it can be broken down into simpler AND operations.

For instance, if we apply the Distributive Law to \((a+b)(c+a+\bar{h})\), we distribute the term \(a+b\) over the term \(c+a+\bar{h}\). Here's how it works in our given problem:

• First, we separate \(a+b\) from the other terms.
• We then distribute \(a\) and \(b\) over each remaining term in the expression.
• Finally, we apply this law repeatedly until no further distribution is possible.

Through persistent application of the Distributive Law, we can turn a tangled web of Boolean logic into a clearer and more manageable set of expressions.

Boolean Algebra Rules

Boolean algebra is governed by a set of rules that help in simplifying expressions through the application of operations such as AND, OR, and NOT. Let's highlight some essential Boolean algebra rules used within our exercise:

• Idempotent Law: This law states \( A + A = A \) and \( A \times A = A \) suggesting that repeating the same variable in an OR or AND operation will not change the outcome, as seen in our exercise where \( aca = ac \).
• Complement Law: It denotes that \( A + \bar{A} = 1 \) and \( A \times \bar{A} = 0 \), explaining why \( acb = ac \times b = ac \) as seen in step 2.
• Zero and Identity Law: These laws denote that \( A + 0 = A \) and \( A \times 1 = A \), highlighting why certain terms are dropped or included.
• Annihilation Law: According to this law, \( A \times 0 = 0 \) which helped in removing certain redundant terms in the earlier steps.

These rules lay the cornerstone for simplifying Boolean expressions. By applying a combination of these rules strategically, we can significantly reduce the complexity of Boolean expressions to a form that is much easier to understand and work with.

Factor Out Common Terms

When simplifying Boolean expressions, 'factoring out' common terms is a vital process that helps to compress the expression into a more concise form. Much like finding a common denominator in algebraic equations, factoring involves identifying and extracting common Boolean terms from an expression.

Through this method, we can avoid redundancy and simplify the logic operation. For example, if we observe our final expression \( ac+\bar{h}a+\bar{h}b\bar{a}+ahcd+\bar{c}+\bar{h} \), we can see \( ac \) and \( \bar{h}a \) as common factors. Upon factoring them out, we streamline the expression, potentially revealing further opportunities for simplification using the Boolean algebra rules mentioned earlier.

By continually factoring out common terms, we’re aiming to reach the simplest form of the Boolean expression. However, it’s crucial to continually assess the simplification at each step to verify the validity of the factorization and ensure it adheres to Boolean algebraic principles.