# Given that U = {a

We know that De Morgan’s laws of contemplation are:

(a) (A B)’ = A’ B’

(b) (A B)’ = A’ B’

Case (a):

First, we shall verify (a). To do this, we consider

A B = {a, b, f, g} {a, b, c} = {a, b, c, f, g}

(A B)’ = U \ {a, b, c, f, g} = {d, e, h} … (1)

Now, A’ = U \ A = {c, d, e, h}

B’ = U \ B = {d, e, f, g, h}

Then, A’ B’ = {c, d, e, h} {d, e, f, g, h} = {d, e, h} … (2)

From (1) and (2), it follows that (A B)’ = A’ B’.

Case (b):

Next, we shall verify (b). To do this, we consider

A B = {a, b, f, g} {a, b, c} = {a, b}

(A B)’ = U \ {a, b} = {c, d, e, f, g, h} … (3)

Now, A’ = U \ A = {c, d, e, h}

B’ = U \ B = {d, e, f, g, h}

Then, A’ B’ = {c, d, e, h} {d, e, f, g, h} = {c, d, e, f, g, h} … (4)

From (3) and (4), it follows that (A B)’ = A’ B’.

Hence, De Morgan’s laws of contemplation are verified.

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