**DeMorgan’s Theorem Statement: **

**The complement of the sum of two or more variables is equal to the product of the complements of the variables.** *If X and Y are the two logical variables, then according to the De Morgan’s Theorem we can write*: **(X + Y)’ = X’.Y’.**

For *n* number of logical variables say *X1, X2, ………………., Xn* it can be written as: *(X1 + X2 + …………+ Xn)’ = X’1.X’2…………….X’n.*

* 2. ***The complement of the product of two or more logical variables is equal to the sum of the complements of the variables.** *If Z and Y are the two logical variables, the according to the second law of De Morgan’s theorem we can write:* **(X.Y)’ = X’ + Y’**

For *n* numbers of variables say *X1, X2, ………………, Xn* the second theorem can be written as: *(X1.X2……………Xn)’ = X’1 + X’2 + ……………….. + X’n.*

**Proof:**

There are two De Morgan’s theorems:

- (X+Y)’ = X’.Y’
- (X.Y)’ = X’+ Y’

First let us prove** (X+Y)’ = X’.Y’** as:

To prove this theorem, we will use complementary laws that are as follows: **X+X’ = 1** and, **X.X’ = 0**.

Now, let P = X+Y, where P, X, Y are three logical variables,

According to the complementary laws we have:

**P+P’ =1 and P.P’ = 0, **if P = X+Y, then P’ = (X+Y)’. If we take (X+Y)’ = X’.Y’.

Therefore to prove (1) we have to show that,

**(X+Y)+(X’.Y’) should be equal to 1** *and* **(X+Y).(X’.Y’) should be equal to 0.**

(X+Y)+(X’.Y’) *[BY X+YZ = (X+Y)(X+Z)]*

= (X+Y+X’).(X+Y+Y’) *[BY X+Y = Y+X]*

= (X+X’+Y).(X+Y+Y’) *[BY X+X’ = 1]*

= (1+Y).(X+1) *[BY X+1 = 1]*

= 1

So, the first part is proved. Now, we have to prove the second part **(X+Y).(X’.Y’)**

= X.X’.Y’ + Y.X’.Y’

[BY X.Y = Y.X]

= X.X’.Y’ + X.Y.Y’

[BY X.X’ = 0]

So, the second part is also proved.

Therefore, we can say that **(X+Y)’ = X’.Y’**

Therefore to prove (2) we have to show that, **De Morgan’s theorem (X.Y)’ = X’+ Y’ **is proved in same way by letting P = X.Y.

Here, we again use the complementary laws.

**X+X’ = 1 and X.X’ = 0. If we take P = X.Y, then P’ = (X.Y)’.**

If we assume that P’ = (X.Y)’ = X’+Y’ then, we have to prove,

(X.Y)+(X’+Y’) should be equal to 1.

(X.Y).(X’+Y’) should be equal to 0.

(X.Y)+(X’+Y’)

[BY X+YZ = (X+Y)(X+Z)]

= (X+X’+Y’).(Y+X’+Y’)

[BY X+Y = Y+X]

= (X+X’+Y’).(X’+Y+Y’)

[BY X+X’ = 1]

= (1+Y’).(X’+1)

[BY 1+X’ = 1]

So, the first part is proved.

Now, we have to prove the second part as:

(X.Y).(X’+Y’)

[BY X.(Y+Z) = X.Y+X.Z]

= X.Y.X’ + X.Y.Y’

[BY X.Y = Y.X]

= X.X’.Y + X.Y.Y’

[BY X.0 = 0]

So, the second part is also proved.

Therefore, we can write** (X.Y)’ = X’+ Y’**

Hence the proof.

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