Hello friends, welcome to get it zil, myself Jayashree Gupta. In this video, we will discuss how to minimize a Boolean expression. Suppose, a Boolean expression is a bar b plus a plus a b bar.
We have written this expression in 3 terms. A bar B is a term, another term is A, another term is A, B bar. So, this is the sum of 3 terms.
Can we express this expression in more complex terms? express or write? Can we further minimize it?
There can be two ways to minimize Boolean expression. Either using the laws of Boolean Algebra or with the help of Cornouche Map. In this video, we are discussing that how to minimize a Boolean expression. To minimize a Boolean expression, we use some laws.
I am discussing the laws that we frequently use for minimizing. So, very first is a plus a bar. How much will it be?
These are the laws which we will use frequently and which we use in minimization. Other than that, next law which we use frequently is Dee Demorgan's law. So, what does Dee Demorgan's law say? So, what is a plus b whole bar?
a plus b whole bar what will be? It is a bar plus get converted into dot or b bar. Complement every literal.
If there is a whole bar of an expression, then to solve that whole bar, we can use Dee Demorgan's law. Dee Demorgan's law says, complement every literal. Replace plus by dot and dot by... by plus. Similarly, if you have a dot in your expression, then you can convert it to a dot.
a plus b plus c bar whole bar. What will happen? What to do with every variable and every literal is complement. So, a bar a is complement of a. This plus will be converted into dot.
Then b complement. Then, what is this plus converted into? Deot. Then, c is the complement of the letter j.
So, c is the complement of the complement. What is the bar of c? It is what? a bar b bar c.
Similarly, if you are asked a bar of a bar, then you will get a bar of a bar. a bar b whole bar. What will happen? What will you do?
Compliment every literal. So, it was already bar. So, you took its bar.
Here, it is dot. So, you converted it into plus. What did you take?
b bar. So, a bar b whole bar is what? a plus b bar.
So, what is the next law? Deemorgan's. After demorgan's, the next law is distributive law. So, what is distributive law? So, distributive law is A dot B plus C is a dot b plus a dot c.
Okay? We have distributed dot on plus. So, if we want, we can take a as common.
If we take common, this expression will be there. And if we distribute a, this expression will be there. expression.
So, in Boolean algebra dot is distributive over plus. In similar way plus is also distributive over dot. So, if your expression is a plus b dot c.
So, what can you write? a plus b dot c is a plus b dot c. So, what can you write? a plus b dot c b dot a plus c both plus and dot satisfy distributive property. Here, we can take a common.
What will a plus b dot c become? We have distributed a. After distributive property, next law is useful is redundant literal rule. What is redundant literal rule? Suppose, we have an expression a plus a bar b.
Here a is there plus in this a is a complementary term. So, we have a plus a bar b. So, we have a plus a bar b. So, we We have a bar.
So, the redundant literal rule says that this a bar is redundant. a plus a bar is something. So, a bar will be redundant. What will be our expression?
It is a plus b. a plus a bar. b is a plus b. So, how did a bar become redundant?
What was our original expression? a plus a bar b. Here, what do we do on the dot of plus?
Deistribute. So, if we distribute plus on dot, then a plus a bar dot a plus b. We have applied this distributive property. Now, what is a plus a bar? 1. So, it is 1 dot a plus b.
So, what is a plus b? So, a plus a bar b. is equivalent to a plus b. Redundant literal. Which literal is redundant?
Extra is a bar. Similarly, if your expression is like this, a bar plus abc. So, here who will be redundant? This is a bar.
In plus, its complementary term is a. So, here a is redundant. You can simply remove this a and write this expression as a bar plus bc. So, we will identify redundant literal.
Here, there is a variable and a product term with plus. What will be the complement of the variable in that product term? Here, there is a bar plus a product term. So, what is the complement of a bar?
a. So, what is present with product term? a.
So, what will be a? redundant, extra. It is extra, so what should we do? Remove it.
So, these are the few laws. There are many laws in Boolean algebra. But these are the few laws which we frequently use in minimization. Which are they? Deistributive law, Dee Demorgan's law, redundant literal law, and the rest that we had written a plus a bar equal to 1 and so on.
So, with the help of these laws, we will minimize some Boolean expressions. I have asked you to minimize this Boolean expression. So, the first thing that we see in minimization is bc and b bar c. Both these terms have c. So, we take common C.
So, if we take common C, then it is A plus B plus B bar. What else did we take? C. B plus B bar is 1. So, it is A plus C. So, this Boolean expression minimizes to what?
A plus C. Now, we will minimize this Boolean expression. In minimization, what do you see in the very first term? What can we take as common in both terms?
a bar in the first and third term. So, we took a bar as common. a bar is common, so it is b plus b bar plus ab.
How much is b plus b bar? 1. That is a bar dot 1 plus ab. How much will a bar dot 1 be simply? a bar. a bar plus a b.
Now, what can you apply in a bar plus a b? Redundant literal rule. Here a is complemented plus a is product term where its complemented form is present.
So, what will a become? Pencil. And ultimately what will be left?
a bar plus b. So, this expression simplifies to a bar plus b. So, next a plus a b bar plus a bar b.
What is common from here? a. Both terms have a as common. So, if we take a as common, it is 1 plus b bar plus a bar b.
How much is 1 plus b bar? So, 1 plus b bar is 1 and we have already studied that 1 plus b bar is equal to 1. So, what is this term? a dot 1 plus a bar b is a. a dot 1 is a. So, a plus a bar b.
Then what will be applied? Our redundant literal rule. Here a and here its complemented term.
So, a cancel or it is simplified to a plus b. So, let us take another example for minimization. We have to minimize this Boolean expression. Expression a b plus a c whole bar plus a b bar c into a b plus c.
So, first of all we will solve this expression. A is open. So, what we can apply here is Dee Demorgan's law.
What does Dee Demorgan's law say? A whole bar is a bar plus c bar. So, it is ab plus a bar plus c bar.
Here, we will product these two terms. So, it is a b bar c ab. Now, we will see these two terms. Here, B bar and B are coming in the product. What is B bar dot B?
It is 0. And, here C is in the product twice. C dot C is simply C. So, this term is reduced to 0. And, this is A B bar C. So, what will happen?
A B plus A bar plus C bar plus 0 plus A B bar C. Okay. Now, how to simplify further?
For further simplification, here is C bar and here is C. These are two contradictory terms. So, we can remove C using redundant.
So, what will be a b plus a bar plus c bar and plus 0 is not important. Something plus 0 is always something. So, we have removed 0 plus a bar plus c bar plus a bar plus a a b bar c.
So, redundant literal rule says that here c bar is there and here c is there so c is redundant. So, we removed c. Now, what will be our term? a b plus a bar plus c bar plus a.
Now, if I take a bar and a, then again what can we apply? Redundant literal rule. Here A bar and A. So what will be A?
Cancel. So, it is a b plus a bar plus b bar plus c bar. Then in both terms you can apply redundant literal rule. Here a bar is here and here a is here.
So, what is a? Redundant. So, it is b plus a bar plus b bar plus c bar.
Now, here what term you get? b plus b bar. How much is b plus b bar? 1. So, it is 1 plus a bar plus c bar and 1 plus something is always 1. So, this expression minimizes to 1. Let us see one more example.
Now, we will simplify this expression. First of all, what is this expression? a b bar plus a b c whole bar plus a into b plus a b bar and what is the whole bar of this expression?
First of all, we will do multiplication in this expression. So, what will be our expression? a b bar plus a b c. If we multiply this, it is A plus A into A is simply A.
What is our expression? What is our expression? A bar plus A plus A plus A bar.
a into a is a and a b bar. Now, what we take common from here? a. So, if we take a common from here, then what will be it?
a b bar plus this will remain as it is plus if we take a common from here, then it is b plus b bar and we know that b plus b bar is 1. So, a dot 1 is simply a. So, what is our expression? a b bar plus a b bar is 1. So, we have a common Now, let us talk about this. What can we take as common in this? A.
If we take A as common from here, then what is our expression? If we take A as common, then it is b bar plus b c and this is plus 1. plus a. Now, what is seen here? This is b bar and here b, two contradictory terms.
So, which rule can we apply? Redundant literal. cancel and this b cancels. If b is cancelled, what will happen?
b bar plus c plus a. Now, let us multiply a. So, it is a b bar plus a c plus a.
Now, to solve this bar, we apply Dee Demorgan's law. So, what will happen if we apply Dee Demorgan's law? Whole bar of first term will be connected to dot and whole bar of second term is same.
So, what will happen? So, this is a b bar whole bar dot. A whole bar and this is plus. This will remain as it is because this is not the bar inside the whole bar.
If we solve this, then what will happen is A, here De Demorgan is applied. That is, this is A. a bar plus b dot plus c bar plus it is a. Now, we will do the product of both terms.
The product of both terms is a bar into a bar is simply a bar plus it is a bar. C bar plus A bar B plus B C bar plus A. Now, in this whole expression, you can see A bar and A.
So, A bar plus A is 1. So, what did we write here? It is 1 plus a bar c bar plus a bar b plus a bar b plus b c bar. Now, 1 plus something we have is always 1. So 1 plus something is always 1. So our expression is only 1 bar.
And 1 bar is 0. So this expression is simplified to 0. So we are simplifying the next expression. A plus B whole bar into A bar plus A. This whole expression is a whole bar and these are the entities.
First of all, here we apply Dee Demorgan's law. B whole bar. If we apply Dee Demorgan's law on B whole bar, then what happens?
B bar plus C bar. So, what is our expression? a plus b bar plus c bar. We have put demorgan on b c bar.
Now, let us see what is common here. If we take a common here, then it is b bar plus b c. Now, here b bar a and b again redundant literal rule b bar a and here b c is b cancel.
Now, we apply the Dee Demorgan's law. We have applied Dee Demorgan's law because there is a whole bar. So, what will happen? We will add these two dots in dot.
And what will happen? a bar dot b complement is simply b dot c complement. This is a term into if I had producted them, that is a b bar plus a c.
Now, what will be our term? a bar bc into ab bar plus ac. Now, if you multiply these terms, then what will be the sum of these terms?
So, what will be the term? Multiply it. a bar bc into ab bar plus a bar bc into ac.
What will be the term? a bar and a. two contradictory terms.
How much is the product of b bar and bn? 0. So, how much is this whole term? 0. Similarly, here also a bar dot a is present. How much is a bar dot a? 0. So, this term is also 0. This term is also 0. So, 0 plus 0. How much is our final sum?