(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 hen 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’