(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 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
(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
