FIRST THEOREM:

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 first law we can write:

$(X + Y)’ = X’.Y’$

SECOND THEOREM:

The complement of the product of two or more logical variables is equal to the sum of the complements of the variables.

If X and Y are the two logical variables, the according to the second law of De Morgan’s we can write:

$(X.Y)’ = X’ + Y’$

PROOF:There are two theorem of DeMorgan’s theorem

1. $(X+Y)’ = X’.Y’$
2. $(X.Y)’ = X’+ Y’ $

PROOF OF FIRST LAW:

First we will prove the first law of the DeMorgan’s theorem.     

1. $(X+Y)’ = X’.Y’$

To prove this theorem, we will use complementarily 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, then according to the complementarily laws we can write:

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 the first law we have to prove that,

(X+Y)+ (X’.Y’) should be equal to 1.

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

$\begin{align*} (X+Y)+ (X’.Y’)&=((X+Y)+X’).((X+Y)+Y’)\\ &[ \because X+YZ = (X+Y)(X+Z)]\\ &=(X+Y+X’).(X+Y+Y’)\\ &= (X+X’+Y). (X+Y+Y’)\\&[\because X+Y = Y+X]\\ &= (1+Y). (X+1)\\&[\because X+X’ = 1]\\ &= 1.1\\&[\because X+1 = 1]\\ &= 1 \end{align*}$

So, the first part is proved.

Now, we have to prove the second part

$\begin{align*} (X+Y). (X’.Y’) &= X.X’.Y’ + Y.X’.Y’\\ &[\because X. (Y+Z) = X.Y+X.Z]\\ &= X.X’.Y’ + X.Y.Y’\\ &[\because X.Y = Y.X]\\ &= 0.Y’ + X.0\\ &[\because X.X’ = 0]\\ &= 0 + 0= 0 \end{align*}$

So, the second part is also proved.

Therefore, we can say that

$(X+Y)’ = X’.Y’$

PROOF OF SECOND LAW:

1. $(X.Y)’ = X’+ Y$

Second law of the Dr Morgan’s theorem is proved in same way by letting P = X.Y.

 Here, we again use the complementarily 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. 

$\begin{align*} (X.Y)+ (X’+Y’) &= (X+X’+Y’).(Y+X’+Y’)\\ &[\because X+YZ = (X+Y)(X+Z)]\\ &= (X+X’+Y’).(X’+Y+Y’)\\ &[\because X+Y = Y+X]\\ &= (1+Y’). (X’+1)\\ &[\because X+X’ = 1]\\ &= 1.1\\ &[\because 1+X’ = 1]\\ &= 1 \end{align*}$

So, the first part is proved.

$\begin{align*} (X.Y).(X’+Y’) &= X.Y.X’ + X.Y.Y’\\ &[\because X.(Y+Z) = X.Y+X.Z]\\ &= 0.Y + X.0\\ &[\because X.0 = 0]\\ &= 0+0= 0 \end{align*}$

So, the second part is also proved.

Therefore, we can write

$(X.Y)’ = X’+ Y’$