**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’**

*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:

**X+X’ = 1**and,

**X.X’ = 0**.

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

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

*and*

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

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

*[BY X+Y = Y+X]*

*[BY X+X’ = 1]*

*[BY X+1 = 1]*

**(X+Y).(X’.Y’)**

[BY X.Y = Y.X]

[BY X.X’ = 0]

**(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)’.**

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

[BY X+Y = Y+X]

[BY X+X’ = 1]

[BY 1+X’ = 1]

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

[BY X.Y = Y.X]

[BY X.0 = 0]

**(X.Y)’ = X’+ Y’**