Boolean Algebra: 10 Solved Exercises
Welcome to our new article, dear followers! In this piece, we will put into practice the concepts we explored in our previous articles on Boolean algebra. You can review those foundational lessons here: Boolean Algebra and Simplification of Logical Algebraic Expressions , Digital Electronics: Laws of Boolean Algebra Here, we will work through practical exercises focused on simplifying Boolean algebraic expressions by applying the fundamental rules and theorems of Boolean algebra.
📚 Boolean Algebra Laws Quick Reference
A + 0 = A
A · 1 = A
A + 1 = 1
A · 0 = 0
A + Ā = 1
A · Ā = 0
A + A = A
A · A = A
A + AB = A
A(A+B) = A
A(B+C) = AB+AC
A+BC = (A+B)(A+C)
ĀB = Ā+B̄
Ā+B = Ā·B̄
AB+ĀC+BC = AB+ĀC
(A+B)(Ā+C)(B+C) = (A+B)(Ā+C)
Exercise 01: Simplify \( F = A + A \cdot B \)
In the first expression, we encounter the Absorption Law, and as we previously studied, the expression simplifies to A. Now, we aim to demonstrate the proof of this result.
First, we identify a common factor in the expression, which is A. Therefore, we will perform factoring to transform the expression into the following form:
This leaves us with the expression (B+1) inside the parentheses. According to the Domination Law (also called the Identity Law for OR), (B+1) simplifies to 1, as any Boolean variable ORed with 1 always results in 1.
Consequently, our expression is reduced to A · 1. Applying the Identity Law for AND, which states that any variable ANDed with 1 equals the variable itself, we obtain the final simplified result:
Exercise 02: Simplify \( F = A \cdot (A + B) \)
Now, let's examine the expression A(A+B). This is a direct application of the Distributive Law of Boolean Algebra.
By distributing the external variable A to each term inside the parentheses, we expand the expression as follows:
Next, in our resulting expression, we have the term A·A. According to the Idempotent Law (which states that a variable ANDed with itself equals the variable), A·A simplifies to A.
Therefore, our expression becomes:
We have already proven in Exercise 01 that the expression A + A·B simplifies to A using the Absorption Law. Thus, we confirm our final result.
Exercise 03: Simplify \( F = (A + B) \cdot (A + \overline{B}) \)
The expression F = (A + B)·(A + B̅) can be simplified using multiple approaches.
First approach: Using the Distributive Law. While valid, this method involves several intermediate steps.
Second (and optimal) approach: Using the Factoring Technique. This is the most efficient and straightforward method. We observe that the common factor in both product terms is the variable A.
After factoring out A, we obtain:
Now, we examine B·B̅. According to the Complement Law (also called Negation Law), a variable ANDed with its complement always equals 0:
Substituting this result back into our expression gives us:
Finally, applying the Identity Law for OR, which states that any variable ORed with 0 equals the variable itself, we obtain:
Exercise 04: Simplify \( F = A \cdot B + A \cdot \overline{B} \)
Let's examine the expression F = A·B + A·B̅. This expression contains two terms with a common factor.
First, we identify that the variable A is present in both terms. We can therefore apply the factoring technique to simplify the expression. We factor A out of both terms:
Now, we focus on the expression inside the parentheses: (B + B̅). This is a direct application of the Complement Law (also known as the Negation Law). The Complement Law states that for any Boolean variable, the expression (variable + its complement) always equals 1. Therefore:
Substituting this result back into our factored expression gives us:
Finally, we apply the Identity Law for AND. This law states that any Boolean variable ANDed with 1 is equal to the variable itself. Consequently:
Exercise 05: Simplify \( F = A + \overline{A} \cdot B \)
Let's examine the expression F = A + Ā·B. We can simplify this expression using algebraic manipulation with Boolean laws.
First, we notice that we can add a term A·B without changing the expression's value, since A + A·B = A by the Absorption Law. This technique is useful for creating common factors:
Now we have three terms. We can group the last two terms together since they both contain B:
Now we examine the expression (Ā + A) inside the parentheses. According to the Complement Law, a variable ORed with its complement always equals 1:
Substituting this result back into our expression gives us:
Now we apply the Identity Law for AND, which states that any variable ANDed with 1 equals the variable itself:
Therefore, our simplified expression becomes:
Exercise 06: Simplify \( S_1 = \overline{A} \cdot B + A \cdot \overline{B} + A \cdot B \)
Let's examine the expression S₁ = Ā·B + A·B̄ + A·B. We have three terms, and we can see that there are opportunities for simplification.
First, we notice that the last two terms A·B̄ and A·B both contain the common factor A. Let's group these two terms together:
Now we examine the expression (B̄ + B) inside the parentheses. According to the Complement Law, a variable ORed with its complement always equals 1:
Substituting this result back into our expression gives us:
Now we apply the Identity Law for AND, which states that any variable ANDed with 1 equals the variable itself:
Now our expression becomes:
We can rearrange the terms (commutative property of OR):
This is exactly the form we simplified in Exercise 05. Using the result from that exercise, we know this simplifies to:
Exercise 07: Simplify \( S_2 = B + A \cdot B + B \cdot C + C \)
Let's examine the expression S₂ = B + A·B + B·C + C. We have four terms, and we can look for opportunities to apply Boolean laws for simplification.
First, we notice that the first two terms B and A·B have a common factor B. Let's factor B from these two terms:
Now we examine the expression (1 + A). According to the Domination Law for OR, anything ORed with 1 equals 1:
Substituting this result back into our expression gives us:
Now we apply the Identity Law for AND, which states that any variable ANDed with 1 equals the variable itself:
Now our expression becomes:
Now we notice that the first two terms B and B·C again have a common factor B. Let's factor B from these two terms:
Again, we examine (1 + C). According to the Domination Law for OR:
Substituting this result gives us:
Applying the Identity Law for AND again:
Therefore, our final simplified expression is:
Exercise 08: Simplify \( ABC + \overline{A} + \overline{C} \)
Let's examine the expression \(F = ABC + \overline{A} + \overline{C}\). We have three terms, and we can look for opportunities to apply Boolean laws for simplification.
First, let's rearrange the terms to group related terms together. We can use the Commutative Law to reorder the terms:
Now, let's focus on the first two terms \(\overline{A} + ABC\). This has a form similar to the Absorption Law. We can factor A from the term ABC:
Using a variation of the Absorption Law (specifically, the form \( X + XY = X + Y \)), we can simplify \(\overline{A} + A \cdot BC\):
Now our expression becomes:
Now let's rearrange again to group BC and \(\overline{C}\) together:
Now, let's examine \(BC + \overline{C}\). We can factor C from the first term:
Using the same variation of the Absorption Law (\( X + XY = X + Y \)) but with complemented form, we can simplify \(C \cdot B + \overline{C}\):
Now substituting this result back into our expression gives us:
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