DeMorgan’s theorems state that

(i) (X + Y)’= X’.Y’

(ii) (X.Y)’= X’ + Y’

(i) (X + Y)’= X’.Y’ 

Now to prove DeMorgan’s first theorem, we will use complementarity laws. Let us assume that P = x + Y where, P, X, Y are logical variables. Then, according to complementation law 

P + P’ =1 and P . P’= 0 

That means, if P, X, Y are Boolean variables then this complementarity law must hold for variables P. In other words, if P i.e., if (X + Y)’= X’.Y’then

(X + Y) + (XY)’must be equal to 1. (as X + X’= 1)

(X + Y) . (XY)’must be equal to 0.

(as X . X’= 0) Let us prove the first part, i.e.,

(X + Y) + (XY)’ = 1

(X + Y) + (XY)’= ((X + Y) +X’).((X + Y) +Y’) (ref. X + YZ = (X + Y)(X + Z))

= (X + X’+ Y).(X + Y +Y’) 

= (1 + Y).(X + 1) (ref. X + X’=1)

= 1.1 (ref. 1 + X =1)

= 1

So first part is proved.

Now let us prove the second part i.e.,

(X + Y) . (XY)’= 0

(X + Y) . (XY)’ = (XY)’ . (X + Y)   (ref. X(YZ) = (XY)Z)

= (XY)’X + (XY)’Y    (ref. X(Y + Z) = XY + XZ)

= X(XY)’ + X’YY’ 

= 0 .Y + X’ . 0     (ref. X . X’=0) 

= 0 + 0 = 0

So, second part is also proved, Thus: X + Y = X’ . Y’

(ii)  (X.Y)’= X’ + Y’ 

Again to prove this theorem, we will make use of complementary law i.e., 

X + X’= 1 and X . X’= 0

If XY’s complement is X + Y then it must be true that

(a) XY + (X’+ Y’) = 1 and (b) XY(X’+ Y’) = 0 

To prove the first part

L.H.S = XY + (X’+Y’)

= (X’+Y’) + XY      (ref. X + Y = Y + X) 

= (X’+Y’ + X).(X’+Y’ + Y)     (ref. (X + Y)(X + Z) = X + YZ)

= (X + X’+Y’).(X’ + Y +Y’) 

= (1 +Y’).(X’ + 1)      (ref. X + X’=1)

= 1.1 (ref. 1 + X =1)

= 1 = R.H.S

Now the second part i.e.,

 XY.(X + Y) = 0

L.H.S = (XY)’.(X’+Y’)

= XYX’ + XYY’ (ref. X(Y + Z) = XY + XZ) 

= XX’Y + XYY’ 

= 0.Y + X.0   (ref. X . X’=0) 

= 0 + 0 = 0 = R.H.S. 

XY.(X’ + Y’)= 0 and XY + (Xʹ +Y’) = 1 

(XY)’= X’ + Y’. Hence proved.