YouTube Transcript:
De Morgan's Law in Boolean Algebra Explained (with Solved Examples)

Skip watching entire videos - get the full transcript, search for keywords, and copy with one click.

Video Transcript

Hey friends, welcome to the YouTube  channel ALL ABOUT ELECTRONICS.

So in the previous video, we have  seen the different Boolean laws.

The only thing which remained was the De  Morgan's theorem or the De Morgan's law.

So in this video, let us understand that.

Because these laws are very useful in minimizing  the Boolean expression or the Boolean functions.

And first, let's see this first law.

So as per this law, the complement of A  plus B is equivalent to A bar dot B bar.

So if we see from the logic gate perspective,  then this A plus B whole bar is equivalent

to the two input OR gate followed by the NOT gate.

So we know that if A and B are the input to  this OR gate, then its output is equal to A + B.

And when it is passed through the NOT gate,  then its output will be equal to A plus B

whole bar.

So basically, this NOT gate inverts the output.

And we also know that this OR gate followed  by the NOT gate is equivalent to the NOR gate.

That means if we apply the A and B input to  this NOR gate, then its output is equal to

A plus B whole bar.

That is equal to the left hand side.

So similarly, let's see the RHS.

So this A bar dot B bar is equivalent to applying  the input A bar and B bar to the AND gate.

And these A bar and B bar signals can be generated  by passing this input A and B through the

NOT gate.

Now, this circuit is often drawn  in the abbreviated form like this.

That is the bubbled AND gate, where this bubble  at the input indicates that the input gets

inverted before the AND operation.

So as per this De Morgan's law, this NOR  gate is equivalent to the bubbled AND gate.

That is A plus B whole bar is  equivalent to the A bar dot B bar.

So using the truth table,  now let's also prove that.

So for the two inputs A and B, there  are total four different possibilities.

So first of all, let's find  this A bar and the B bar.

So this A bar is the complement of the A.  That means when A is 0, then this A bar is

equal to 1.

And likewise, when this A is equal  to 1, then this A bar is equal to 0.

Similarly, this B bar is the complement of the B.

So once we have this A bar and the B bar,  then we can easily find this A bar dot B bar.

So this A bar dot B bar is 1, when  both inputs A bar and B bar are 1.

And when any of them is 0,  then this output is equal to 0.

So in this way, we got this A bar dot B bar.

Alright, so now let's find this A plus B.  So this A plus B is 0, when both inputs A

and B are 0.

And when any of them is 1, then  this A plus B is equal to 1.

Now this A plus B whole bar is  the complement of this A plus B.

So in this way, we also got  this A plus B whole bar.

And if you see these two columns, then we  can say that this A bar dot B bar and this

A plus B whole bar are same.

So this law is valid even  for more than two inputs.

That means if we have three inputs A, B and  C, then this A plus B plus C whole bar is

equivalent to the A bar dot B bar dot C bar.

Likewise, for the four inputs, this A plus  B plus C plus D whole bar is equivalent to

the A bar dot B bar dot C bar dot D bar.

So that is the De Morgan's first law.

Similarly, let's see the second law.

So as per this second law, the complement  of the A dot B is same as the A bar plus B

bar.

So first of all, let's see the same  from the logic gate perspective.

So this A dot B whole bar is equivalent  to the AND gate followed by the NOT gate.

So the output of this AND gate will be equal  to A dot B. And this NOT gate inverts that

output.

That means the overall output of this  combination is equal to A dot B whole bar.

Now we know that this AND gate followed by  the NOT gate is equivalent to the two input

NAND gate.

That means if we apply the input A and B to  the NAND gate, then its output is equal to

A dot B whole bar.

That is equal to the left hand side.

Similarly, let's see the RHS.

So this A bar plus B bar is equivalent to  applying the A bar and B bar input to the

OR gate.

And this A bar and B bar signals can be generated  by passing the inputs A and B through the

NOT gate.

And as we have seen, this circuit is equivalent  to the bubbled OR gate, where this bubble

inverts the input before the OR operation.

That means as per this second De Morgan's law,  the NAND gate is equivalent to the bubbled

OR gate.

So now, let's prove the  same using the truth table.

And first of all, let's find  this A bar and the B bar.

So as we have seen, this A bar and B bar  are the complement of the inputs A and B.

So once we have this A bar and B bar, then  we can easily find this A bar plus B bar.

So this A bar plus B bar is  low, when both inputs are low.

And when any of the input is high,  then this A bar plus B bar is high.

So in this way, we got this A bar plus B bar.

Similarly, to find this A dot B whole bar,  first of all, let's find this A dot B.

So this A dot B is high, when  both inputs A and B are high.

And when any of them is zero, then this  output A dot B will be equal to zero.

And this A dot B whole bar  is the complement of this AB.

So now, if you see these two columns, then  we can say that this A dot B whole bar is

equivalent to the A bar plus B bar.

So this law is also valid  for more than two inputs.

That means for the three inputs, we can say  that this A dot B dot C whole bar is equivalent

to the A bar plus B bar plus C bar.

Similarly, for the four inputs, this A dot  B dot C dot D whole bar is equivalent to the

A bar plus B bar plus C bar plus D bar.

So that is De Morgan's second law.

So now these two laws are very useful  in minimizing the Boolean expressions.

So basically, in these two laws, what we are  doing, we are breaking this negation or the

bar, and we are giving it to the  individual variable or the group.

And at the same time, we  are also changing the sign.

That means here, we have an OR operation, and  it will get converted into the AND operation.

And the same thing is happening in the second law.

So basically, we are breaking this bar, and  we are giving it to the individual variable

or the group underneath.

And then, we are also changing this sign.

That means here, this AND operation  gets converted into the OR operation.

And the same rule is also applicable to the  Boolean expressions, which has this sort of

negation.

So let's take a few examples related to Boolean  algebra where we can use this De Morgan's law.

So this is our first example.

So as you can see, this  expression has a bar on top of it.

So what we will do, we will break this bar,  and we will give it to the individual variable

or the group.

And then, we will change the sign.

So first of all, let's break this bar  and give it to the individual group.

That means this A will become A bar, and  this B bar C will have a bar on top of it.

Moreover, we will also change this sign.

That means now, this OR operation will  get converted into the AND operation.

That means now, we will have this  A bar dot this B bar C whole bar.

So once again, now let's break this bar.

So now, we will have this A bar dot B  double bar and this C will become C bar.

Moreover, this AND operation between this  B bar and C will get converted into the OR

operation.

That means now, we will have this A bar  dot this B double bar or D plus C bar.

Because we know that the complement  of the A bar is equal to A.

That means this B double  bar is equivalent to the B.

So in this way, this A plus B bar dot C  whole bar will become A bar dot B plus C bar.

So in this way, using the De Morgan's  law, we can simplify the expression.

Similarly, let's take the second example.

So in this example also, we  will follow the same procedure.

That means first, we will break this bar  and we will give it to the individual group.

That means this A plus C can be  written as this A plus C whole bar.

Likewise, this A plus B bar can be  written as A plus B bar whole bar.

Likewise, we will have this A  bar plus B plus C bar whole bar.

Moreover, these AND operations will  get converted into the OR operations.

Now once again, let's break this bar and  let's give it to the individual variable.

So we can write this A plus C  whole bar as A bar dot C bar.

So basically, we have given this bar to the  individual variable and this OR operation

got converted into the AND operation.

Likewise, this A plus B bar whole bar can be  written as A bar dot B because this B double

bar is equal to B. And likewise,  this A bar plus B plus C bar

whole bar can be written as A double  bar dot B bar dot C double bar.

So once again, we have given this bar to the  individual variable and we have also changed

this sign.

That means now we can write this expression  as A bar dot C bar plus A bar dot B plus A

dot B bar dot C. So in this  way, using this De Morgan's law,

we can simplify the expression.

So similarly, let's take one more example.

So in this example, using this De Morgan's  law, let's simplify this Boolean expression.

So as you can see over here, this expression  in the bracket has a bar on top of it.

So in this bracket, we will apply the De Morgan's  law and then we will simplify this expression.

So first of all, let's write down  the remaining expression as it is.

That is equal to A dot B plus C bar.

Now in the bracket, what we will do, we will  give the bar to each group and then we will

change the sign.

That means we can write this bracket as A  dot B whole bar dot complement of the AC bar.

Now once again, here we can  apply the De Morgan's law.

So we can write this expression as A dot B  plus C bar and this A dot B whole bar can

be written as A bar plus B bar.

Likewise, the complement of the AC  bar can be written as A bar plus C.

So basically, we have broken this bar and  we had given to the individual variable.

And at the same time, we  have also changed the sign.

So as you can see, further we  cannot apply the De Morgan's law.

But using the other Boolean algebra  laws, we can simplify this expression.

So let's do that.

So now we can write this  expression as A dot B plus C bar.

And in the bracket, we will have four terms.

That is A bar dot A bar plus A bar dot  C plus B bar dot A bar plus B bar dot C.

So we know that this A bar  dot A bar is equal to A bar.

Moreover, this B bar dot A bar  is equivalent to A bar dot B bar.

So now we can write this  expression as A dot B plus C bar.

And in the bracket, we will have A bar plus  A bar C plus A bar dot B bar plus B bar dot

So now in these three terms,  we can take this A bar common.

So now we can write this  expression as A dot B plus C bar.

And in the bracket, when we take this A bar  common, then we will have 1 plus C plus B

bar.

And this B bar plus C will remain as it is.

Now we know that 1 plus something is equal to 1.

That means this entire term will be equal to 1.

So now we can write this  expression as A dot B plus C bar.

And in the bracket, this A  bar dot 1 plus B bar dot C.

So further if we simplify it, then we  can write it as A dot B plus C bar.

And in the bracket, we will  have A bar plus B bar dot C.

So now using the distributive law, we  can further simplify this expression.

That is equal to A dot B plus this C bar  dot A bar plus B bar dot C dot C bar.

Now we know that this A dot A bar is equal to 0.

Similarly over here, this  C dot C bar will become 0.

That means this term will become 0.

That means now we will have this  A dot B plus A bar dot C bar.

And once again, using the distributive law,  we can write it as A dot B plus A dot A bar

dot C bar.

And once again, this A dot A bar will become 0.

That means this term will also become 0.

And at the end, we will have AB.

That means this entire term is equivalent to AB.

So in this way, using the De Morgan's law,  we can simplify the Boolean expression.

Alright, so this is the fourth  example for your exercise.

So using the De Morgan's law,  try to simplify this expression.

And do let me know your answer in the comments.

But I hope in this video, you  understood about the De Morgan's law.

So if you have any question or suggestion, then  do let me know here in the comment section below.

If you like this video, hit the like button and  subscribe to the channel for more such videos.

Click on any text or timestamp to jump to that moment in the video

Most transcripts ready in under 5 seconds

One-Click Copy125+ LanguagesSearch ContentJump to Timestamps

Paste YouTube URL

Enter any YouTube video link to get the full transcript

Most transcripts ready in under 5 seconds

Get Our Chrome Extension

Get transcripts instantly without leaving YouTube. Install our Chrome extension for one-click access to any video's transcript directly on the watch page.

Add to Chrome — Free

Works with YouTube, Coursera, Udemy and more educational platforms

Get Instant Transcripts: Just Edit the Domain in Your Address Bar!

YouTube

←

→

↻

https://www.youtube.com/watch?v=UF8uR6Z6KLc

YoutubeToText

←

→

↻

https://youtubetotext.net/watch?v=UF8uR6Z6KLc

YouTube Transcript:De Morgan's Law in Boolean Algebra Explained (with Solved Examples)

Video Transcript

Paste YouTube URL

Transcript Extraction Form

Get Our Chrome Extension

Get Instant Transcripts: Just Edit the Domain in Your Address Bar!

YouTube Transcript:
De Morgan's Law in Boolean Algebra Explained (with Solved Examples)