Question

Find the minimal dnf for the following Boolean function: $$ f(x, y, z)=x+x y(z+x)(\bar{y}+z)+(\underline{x}+y+z) \bar{x}+y z $$

Short answer

The minimal DNF for the given Boolean function is: \(f(x, y, z)=x+y\bar{x}+xy+z\bar{x}+yz\).

Step-by-step solution

Simplify the given Boolean function

We have the following Boolean function: \(f(x, y, z)=x+xyz(x+y)+(\underline{x}+y+z)\bar{x}+yz\) First, let's take care of the four terms: Term1: \(x\) Term2: \(xyz(x+y)\) - Apply the distribution property: \(xyx + xyy\) - Simplify using the idempotent law: \(x*x=x\) and \(y*y=y\) \(x^2y^2 = xy\) Term3: \((\underline{x}+y+z)\bar{x}\) - Apply the distribution property: \(\bar{x}\underline{x} + y\bar{x} + z\bar{x}\) - Simplify using the complement law: \(x*\bar{x}=0\) \(0 + y\bar{x} + z\bar{x}\) - The simplified term becomes: \(y\bar{x} + z\bar{x}\) Term4: \(yz\) Now, we rewrite the Boolean function with the simplified terms: \(f(x, y, z)=x+xy+y\bar{x}+z\bar{x}+yz\)

Simplify the Boolean function further and convert to DNF form

Rewrite the simplified function by collecting the terms together with the x variable as common parts: \(f(x, y, z)=(x+y\bar{x})+(xy+z\bar{x}+yz)\) Now, apply the distribution property in the first part of the function: \(x(1+\bar{x})+y\bar{x}\) Notice that, as per the complement law and the identity law, we have: \(1+\bar{x}=1\) So the first part will be simplified to: \(x+y\bar{x}\) We now have the minimal DNF of the given Boolean function after simplifying it: \(f(x, y, z)=x+y\bar{x}+xy+z\bar{x}+yz\)

Key concepts

Boolean Algebra Simplification

When it comes to understanding Boolean algebra simplification, imagine you are working with the on-off logic of light switches, where 1 represents 'on' and 0 represents 'off'. Simplification in Boolean algebra is similar to solving puzzles: you want to find the simplest way to express the same state of lights using fewer switches. Just like puzzles, there are rules that make this process easier. Simplifying Boolean expressions makes them more comprehensible and can lead to more efficient circuit designs.

In our exercise, simplification helped us reduce the complexity of the function. We began with a complex expression and systematically broke it down to its simplest form without changing its meaning by applying various Boolean algebra rules, like the distribution, idempotent, and complement laws. Put simply, our goal is to turn a confusing tangle of switches (variables and expressions) into a neat and minimal set that controls the lights just as effectively.

Distribution Property

The distribution property in Boolean algebra is akin to breaking down a task into smaller, more manageable parts. It's like dividing a large pizza among friends: instead of giving each friend a specific mix of toppings on a single slice, you serve each topping separately so they can choose exactly what they want on their slice.

Application in the Exercise

In our original problem, we used the distribution property to expand and simplify the term \(xyz(x+y)\). We distributed \(xy\) over \(x+y\), which allows us to look at the effect of each part separately. This application of the distribution law moves us closer to our goal of simplification. In the real world, this translates to optimizing circuits by first dealing with grouped components and then individual ones.

Idempotent Law

The idempotent law in Boolean algebra is a reflection of a fundamental truth: if something is true, repeating it doesn't make it 'more true'. It's like saying 'I am here' once or a hundred times; either way, it's clear that you are present.

Application in the Exercise

In the step by step solution, we simplify \(x*x\) and \(y*y\) to \(x\) and \(y\), respectively. This simplification follows the idempotent law. Essentially, we are peeling away redundancies, like confirming you've heard someone the first time, without needing them to repeat it again and again. This law is important because it helps eliminate unnecessary complexity from Boolean expressions, leaving us with a clearer picture of what's really happening in the system.

Complement Law

The complement law in Boolean algebra states that a thing and its opposite cannot both be true at the same time. It's a bit like saying you can't be both completely wet and completely dry simultaneously. In Boolean terms, a variable and its negation, like \(x\) and \(\bar{x}\), when combined, will always result in 0 (false).

Application in the Exercise

We employed the complement law when looking at \(\bar{x}\underline{x}\) within the term \(\underline{x}+y+z)\bar{x}\). According to this law, the product was simplified to 0, effectively eliminating it from the expression. Complement law comes in handy by allowing us to cancel out certain parts of a Boolean expression, similar to subtracting an amount and its negative in a bank account, ultimately leaving your balance unchanged.