Question

Simplify the following boolean expressions: (a) \((x \wedge y) \vee(x \wedge \neg y) \vee(\neg x \wedge y) \vee(\neg x \wedge \neg y)\) (b) \((x \wedge y \wedge z) \vee(x \wedge \neg y \wedge z) \vee(\neg x \wedge y \wedge \neg z) \vee(\neg x \wedge \neg y \wedge z)\) (c) \((x \wedge y \wedge \neg z) \vee(x \wedge \neg y \wedge z) \vee(x \wedge \neg y \wedge \neg z)\)

Short answer

(a) 1; (b) Complex, can't simplify further easily; (c) \(x \wedge (\neg z \vee \neg y)\).

Step-by-step solution

Simplify Expression (a) using Identities

Consider the expression \[(x \wedge y) \vee (x \wedge eg y) \vee (eg x \wedge y) \vee (eg x \wedge eg y)\]Apply the Distributive Law for the first two terms: \[x \wedge (y \vee eg y)\].By the Law of Excluded Middle, \[y \vee eg y = 1\],therefore \[x \wedge 1 = x\].Similarly, for the last two terms: \[eg x \wedge (y \vee eg y) \Rightarrow eg x \wedge 1 = eg x\].Thus, the simplified expression is \[x \vee eg x\].Finally, apply the Law of Excluded Middle again: \[x \vee eg x = 1\].

Simplify Expression (b) using Factoring Techniques

Consider the expression\[(x \wedge y \wedge z) \vee (x \wedge eg y \wedge z) \vee (eg x \wedge y \wedge eg z) \vee (eg x \wedge eg y \wedge z)\]Group the first two terms using factoring:\[x \wedge (y \wedge z \vee eg y \wedge z)\]Factor out \(z\):\[x \wedge z \wedge (y \vee eg y)\]Since \(y \vee eg y = 1\), the expression becomes \(x \wedge z\).Group the remaining terms:\[eg x \wedge (y \wedge eg z \vee eg y \wedge z)\]This cannot be reduced as easily, so we won't further simplify it right now.Combining both parts, the expression remains the same due to complexity:\[(x \wedge z) \vee (eg x \wedge (y \wedge eg z) \vee (eg y \wedge z))\].

Simplify Expression (c) by Distributing and Factoring

Consider the expression \[(x \wedge y \wedge eg z) \vee (x \wedge eg y \wedge z) \vee (x \wedge eg y \wedge eg z)\]Notice that \(x\) can be factored:\[x \wedge ((y \wedge eg z) \vee (eg y \wedge z) \vee (eg y \wedge eg z))\]Focus on simplifying the internal expression. Notice:\[eg y\] is a common factor for the second and third terms, so it can be factored:\[x \wedge ((y \wedge eg z) \vee eg y)\]Use distribution over first two terms:\[x \wedge (y \vee eg y) \wedge eg z \vee x \wedge eg y\]Since \(y \vee eg y = 1\), the expression simplifies to:\[x \wedge 1 \wedge eg z \vee x \wedge eg y\]which simplifies further to:\[x \wedge (eg z \vee eg y)\]

Key concepts

Boolean Expressions

A Boolean expression is a statement that results in a Boolean value, that is, either true or false. These expressions are composed using Boolean variables and operators such as AND ( \( \wedge \)), OR ( \( \vee \)), and NOT ( \( eg \)). For instance, in Boolean algebra, an expression like \((x \wedge y) \vee z\) indicates a condition where both \(x\) and \(y\) are true or \(z\) is true.

Boolean expressions are fundamental in computer science, often used in decision-making processes within algorithms and programming. Understanding and simplifying these expressions allow for more efficient code execution by reducing complexity or unnecessary computation steps.

• Variables: Represented by letters like \(x, y, z\), where each can be true (1) or false (0).
• Operations: Utilize AND, OR, NOT to form compound expressions.
• Outcomes: These expressions evaluate to a single Boolean value.

By converting complex Boolean expressions into simpler ones, we can ensure systems and circuits perform optimally.

Simplification Techniques

Simplifying Boolean expressions involves applying various algebraic techniques to reduce their complexity while maintaining equivalence. These techniques often make use of known identities and laws pertinent to Boolean algebra, such as:

• Identity Law: \(x \vee 0 = x\) and \(x \wedge 1 = x\).
• Null Law: \(x \vee 1 = 1\) and \(x \wedge 0 = 0\).
• Complement Law: \(x \vee eg x = 1\) and \(x \wedge eg x = 0\).

These techniques allow us to systematically transform complex expressions into simpler, more manageable forms.
For example, consider simplifying expression (a): \((x \wedge y) \vee (x \wedge eg y) \vee (eg x \wedge y) \vee (eg x \wedge eg y)\).
Using the Distributive Law and the Law of Excluded Middle, we observe that \((y \vee eg y)\) is always true (1), reducing the entire expression to 1.
Simplification not only aids in reducing the size of digital logic circuits but also enhances the efficiency of boolean operations in software.

Distributive Law

The Distributive Law is a cornerstone of both traditional and Boolean algebra and plays a critical role in simplifying Boolean expressions. It shows how variables can distribute over operations, similar to how we deal with numbers.

In Boolean algebra, it takes the form: \[ A \wedge (B \vee C) = (A \wedge B) \vee (A \wedge C) \] or the dual: \[ A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C) \].
These equations allow complex expressions to be rewritten in a simpler or more meaningful way.
In the first problem, we utilized this law to group and factor terms effectively. For instance, \( x \wedge (y \vee eg y)\) was formed by distributing \( x \) over the terms inside the parenthesis, showing that \( y \vee eg y = 1 \). This process helped us to substantially reduce the equation to a simpler form: \( x \vee eg x = 1 \).
Understanding the distribution of terms is essential to mastering Boolean algebra transformations, especially in digital logic circuit design.

Law of Excluded Middle

The Law of Excluded Middle is a fundamental concept in logic and Boolean algebra. It states that for any Boolean variable, either the variable itself (\(x\)) or its negation (\(eg x\)) must be true, i.e., \(x \vee eg x = 1\).

This law asserts that there are no other possibilities or 'in-between' values for a Boolean variable, emphasizing the binary nature of Boolean logic. It is an assumption widely used in logic problems, simplifying expressions, and truth table evaluations.

When simplifying Boolean expressions, such as in our exercises, this law came into play. For expression (a), applying this rule by substituting \(y \vee eg y\) with 1 helped condense the expression significantly. Consequentially, any term multiplied by this result slims down to the primary variable or its complement, respectively.
Understanding and applying the Law of Excluded Middle helps effectively minimize Boolean expressions, which is incredibly useful within computational logic and reasoning.