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The Laws of Boolean Algebra Explained

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

So, in the previous video, we learned about the logic gates.

And we have also seen that, when we combine these different logic gates, then it also

performs various logic operations.

So, these logic operations, which are performed by this circuit can be represented using the

Boolean expressions.

Or other way around, if we have some Boolean expression, then it can also be implemented

using these logic gates.

So, in digital electronics, when we use these logic gates, then actually we are also dealing

with these Boolean expressions.

And therefore, it is good to know about some Boolean Laws.

Because using these Boolean Laws, it is possible to minimize the Boolean expression.

And if we can reduce the Boolean expression, then the number of gates which is required

to implement that expression will also reduce.

For example, if we know the Boolean Algebra, then this Boolean expression is equivalent

to A + BC.

So, that is why, it is good to know about the Boolean Laws.

So, here is the list of some Boolean Laws.

So, most of them are pretty straightforward to understand.

And all these laws can be proved using the Truth Table.

So, one by one, let's understand them.

So, the first one is, A.0 = 0 That means the AND operation of this A and

0 is equal to 0.

So, we already know about the two-input AND gate.

And here is the truth table of the two-input AND gate.

So, as we know, for the AND gate, when one of the inputs is 0 then the output is also

zero.

So, in this case, this input B is equal to 0.

So, in this case, since one of the inputs is zero so the output will also become 0.

That means the AND operation of A and 0 is equal to 0.

Then the second is, A.1 = A. That means when we do the AND operation and

the logic '1' then the output is equal to A.

So, it is quite similar to our normal algebra.

But to understand it more clearly, once again let's see the truth table of the AND gate.

So, in this case, one of the inputs is always set to 1.

So, here for the second input A, we have only two possibilities.

So, when A = 0, then the output of this AND gate is equal to 0.

And when A= 1, then both inputs are 1.

That means in that case, the output will be equal to 1.

So, as you can see, this Y column is same as the A.

And therefore, we can say that, this A.1 = A. Then the next is, the AND operation of A and

A is equal to A. that means during the AND operation, when

both inputs are the same, then output is the same as the input.

So, here is the truth table of the two-input AND gate.

And whenever both inputs are the same, then this B is also equal to A.

So, in this case, for the input A, there are two possibilities.

So, when this A is equal to 0, then both inputs are zero.

And in that case, the output of the AND gate will be equal to 0.

And when A= 1, then both inputs are 1.

Therefore, the output of the AND gate will be equal to 1.

So, as you can see, this output Y is following the input.

That means this Y = A. And this is also true, when there are more

than 2 inputs.

For example, if we have A.A.A, then that is also equal to A.

Then the next is the AND operation of A and A-bar is equal to 0.

Where this A-bar is the complement of the A.

So, to prove that, once again let's see the truth table of the AND gate.

So, in this case, this second input B is equal to A-bar.

So, once again, for the input A, there are two possibilities.

That means whenever, A is zero, then this A-bar will be equal to 1.

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

Because this A-bar is the complement of the A.

So, as you can see, in any case, one of the inputs to the AND gate is always zero.

And therefore, the output of the AND gate will always remain zero.

That means this Y is equal to 0.

And that is why, the AND operation of the A and A-bar is equal to 0.

So, these were some of the Boolean laws related to AND operation.

Similarly, let's see the Boolean laws related to OR operation.

So, the first is, A + 0 = 0.

That means the OR operation of A and logic 0 is equal to A.

And this expression is quite similar to our normal algebra.

So, to prove that, once again let's see the truth table of the two-input OR gate.

So, for OR operation, we know that, when one of the input is 1, then the output is always

equal to 1.

And when both inputs are zero, then and then only, the output of the OR gate is equal to

So, here one of the inputs is always set to 0.

So, now for the input A, there are only two possibilities.

So, when this A=0, then both inputs are 0.

And in that case, this output Y is equal to 0.

And when this A=1, then the output of this OR gate is equal to 1.

So, as you can see, this output Y is same as the A.

So, we can say that, this Y=A. That means this A + 0 = A.

Then the next is, A +1 = 1 That means the OR operation of A and the logic

'1' is always equal to 1.

So, once again to prove that, let's see the truth table of the two-input OR gate.

So, in this case, one of the inputs is always set to 1.

So, for the OR gate, we know that, when one of the inputs is 1, then the output is always

equal to 1.

And therefore, irrespective of the value of A, the output Y will always remain 1.

That means this, A + 1 = 1.

Then the next is, A + A = A. That means during the OR operation, if both

inputs are same, then the output is same as the input.

And it is even valid for more than two inputs.

So, to prove that, once again let's see the truth table of the two-input OR gate.

So, in this case, this B is also equal to A.

So, here for the input A, there are only two possibilities.

So, when the input A is 0, then both inputs are 0.

And in that case, this output Y is equal to 0.

And when the input A is equal to 1, then both inputs are 1.

And in that case, this output ty is equal to 1.

So, as you can see, the output Y is same as the input A.

And therefore, this Y = A. That means this A + A = A.

And as I said, it is also valid even for more than 2 inputs.

That means if we have this A + A + A, then that is also equal to A.

Then the next is, A + A-bar is equal to 1.

That means the OR operation of Am with its complement is equal to 1.

So, once again, to prove this expression, let's see the truth table of the two-input

OR gate.

So, in this case, this second input is equal to A-bar.

That is the complement of the input A. So, here for the input A, there are only two

possibilities.

So, when the A is 0, then this A-bar is equal to 1.

And whenever, A is 1, then A-bar is equal to 0.

So, in both cases, at least one input is 1.

And we know that, for the OR gate, when one of the inputs is 1, then the output is equal

That means in both cases, this output Y is equal to 1.

And therefore, we can say that, for this expression, this Y is equal to 1.

Then the next is the commutative law, which says that, the order in which the inputs are

connected to AND gate doesn't matter.

For example, if we connect this input B at the top, and this A is connected to the bottom,

then also, we will get the same result.

That means this A.B is same as the B.A.

And the same is true for the OR gate.

That means if B is connected at the top and A is connected at the bottom, then also, the

output of the OR gate will remain same.

So , that is the commutative law.

Then the next is associative law.

So, as per this law, the AND operation of A and B followed by the C, is same as the

AND operation of B and C with A. So, let's say, the A and B are ANDed first,

and then the output is ANDed with the C. Let's say, this output is equal to Y.

Now, we will get the same result, if B and C are ANDed first, and then its output is

ANDed with the A. That means the AND operation of A and B with

C, is same as the AND operation of B and C with A.

And this can be proved with the help of the truth table.

So, let's say, the output of this A and B, is equal to P.

And the output of the B and C is equal to Q.

So, this P will be high, when both inputs A and B are high.

So, now this P and C are ANDed.

So, its output will be high, when both P and C are high.

And that is happening only in this last possibility.

So, this is the output column Y.

Similarly, now let's find the Y for the second circuit.

And first, let's find this Q.

So, this Q is high, when both B and C inputs are high.

So, now this Q is ANDed with A. And once again, the AND operation of A and

Q is high, when both inputs A and Q are high.

And that is happening only in the last possibility.

So, for the second circuit also this output column Y remains the same, which proves that

both expressions are same.

Similarly, this associative law is also valid for the OR gate.

That means the (A OR B ), ORed with the C is same as the (B OR C) ORed with A.

And it can be also proved with the help of the truth table.

So, let's say, the output of this (A OR B) is equal to P and the output of this ( B OR

C) is equal to Q.

So now, this P is equal to 0 when both inputs A and B are 0.

Otherwise, it is equal to 1.

So, this P ORed with C. So, once again, this output Y is equal to

0 when both inputs P and C are zero.

Otherwise, the output Y is equal to 1.

So, this is the output column Y.

Similarly , let's evaluate this Q for the second circuit.

So, as you can see, this Q is the OR operation of this B and C.

That means this Q is high, when both inputs B and C are high.

And it is 0 when both inputs B and C are zero.

Now, this Q is ORed with the A. And its output is equal to Y.

So, this output Y is equal to 0 when both inputs A and Q are 0.

And when any of the inputs is high then the output Y is equal to high.

So, as you can see, for the both expressions, or for both circuits, this Y column is same,

which proves that both expressions are the same.

So, that is the associative law for the AND and OR operation.

Similarly, it is also true for the XOR and the XNOR operation.

That means the XOR operation of the A and B with the C is the same as the XOR operation

of B and C with A. And the same is also true for the XNOR operation.

And in fact, in the previous video of the XOR and the XNOR gate, we have already seen

that.

That means this XOR and the XNOR gates are also associative.

Now, if we talk about the NAND and the NOR gate, then they are commutative but they are

not associative.

That means in the NAND and NOR gate if we interchange the two inputs then also we will

get the same result.

But they are not associative.

For example, for the NAND gate, the NAND operation of A and B, followed by C is not the same

as the NAND operation of B and C followed by A.

Likewise, for the NOR gate also, the NOR operation on A and B followed by the C is not the same

as the NOR operation of B and C followed by A.

That means these NAND and NOR gates are commutative but they are not associative.

And this can be proved using the truth table.

Alright so now let's move to the next law, which is the distributive law.

So, as per this law, the AND operation of this A and( B + C) is the same as the OR operation

of this AB and AC.

So, basically, this law is very similar to our normal algebra, where the variable outside

the bracket is multiplied with the each term inside the bracket.

So, here the variable outside the bracket gets ANDed with the each term inside the bracket.

Now, this law is also true for more than one variable.

For example, the AND operation of this AB with (C +D) is the same as the OR operation

of this ABC and the ABD.

That means the variable outside the brackets gets ANDed with the each term inside the bracket.

Or if we think other way around, then anything which is common in both terms can also be

taken outside as a common term.

For example, this A + AB is equal to A. So, in this case, this A is common in both

terms.

So, if we take this A outside then we will have A ( 1 + B).

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

That means this term will become 1.

That means now, we will have A.1.

And that is equal to A. That means this A + AB = A.

Similarly, the second distributive law is, A + BC = (A+B) (A+C).

So, to prove that, let's take the RHS.

So, similar to our usual algebra, the each variable gets ANDed with the other variable.

So, in this case, we can write it as A.A + A.C + B.A + B.C

Now, we know that, A.A or the AND operation of A and A is equal to A.

And this B. A can also be written as A.B.

That means we can write this expression as A + AC + AB + BC.

So, in these first three terms, this A is common.

So, if we take this A outside, then we will have 1 + B +C. And this BC will remain as

it is.

Now, once again we know that, 1 plus something is always equal to 1.

That means this is equal to 1.

So, we can write it as A.1 + BC.

And we know that, A.1 = A. That is equal to A + BC.

And that is equal to LHS.

So, that is the second second distributive law.

So, using the same Distributive law we can prove that, this A.(A+B) = A.

Or in other words, the AND operation of A with A+B is equal to A.

So, to prove that, let's take the LHS.

So, here this A will get ANDed with each term inside the bracket.

That means we can write this expression as A.A + A.B.

Now as we have seen earlier, this A.A = A. That means we can write this expression as

A + AB.

And we have just seen that, this A + AB = A. That means the A. (A+B) = A.

So, this is how using these Boolean laws, it is possible to minimize the Boolean expression.

So, in the end, now let's look at the double negation law.

So, we know that, the complement of the A is equal to A-bar.

And if we once again take the complement of the A-bar then it will be equal to A double

bar.

So, if A is equal to 1, then its complement A-bar will be equal to 0.

And once again if we take the complement of this A-bar, then it will be equal to 1.

So, if you see this A double bar, then it is the same as the A.

That means the complement of the A-bar is the same as the A.

So, in this way, we have covered most of the Boolean laws.

Now, the only thing which still remains is De Morgan's Theorem.

So, in the next video, we will also learn about De Morgan's Theorem.

But I hope in this video, you learned bout the different Boolean Laws and their importance

in the digital electronics.

So, if you have any questions or suggestions 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.

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YouTube Transcript:
The Laws of Boolean Algebra Explained