Answer :

__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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