Question

State and prove D'Morgans Law?

Answer 1

Hey

Here is your answer,

Statement:
according to the De Morgan’s law when the complement of the union of given two sets is taken then it is equal to the intersection of the complements of the given sets and when the compliment of the intersection of given two sets is taken then it is equal to the union of the complements of the given sets.

Proof:

We will prove that (X ∩ Y)' = X' ∪ Y'and the other can be proved similarly:

Now to prove this we will prove two things:

i) (X ∩ Y)' ⊆ X' ∪ Y'
ii) X' ∪ Y' ⊆ (X ∩ Y)'

Suppose y ϵ (X ∩ Y)'

→ y ∉ X ∩ Y
→ y ∉ X or y ∉ Y
→ y ϵ X' or y ϵ N'
→ y ϵ X' ∪ Y'

Since there exist a y ϵ (X ∩ Y)' such that y ϵ X' ∪ Y'

→ (X ∩ Y)' ⊆ X' ∪ Y'....(i)

Now consider z ϵ X' ∪ Y'

→ z ϵ X' or z ϵ Y'
→ z ∉ X or z ∉ Y
→ z ∉ X ∩ Y
→ z ϵ (X ∩ Y)'

Since there exist a z ϵ X' ⊆ Y' such that z ϵ (X ∩ Y)'

→ X' ∪ Y'⊆ (X ∩ Y)'...(ii)

From (i) and (ii) we get that

(X ∩ Y)' = X' ∪ Y' that is the De Morgan's law for intersection.

Similarly we can prove the De Morgan's law of union as well.

When we are given more than two sets we get generalize our laws as below:

[∪ni=1(Ai)]' = ∩ni=1 [(Ai)′]

[∩ni=1(Ai)]' = ∪ni=1 [(Ai)′]

Hope it helps you!

Answer 2

In propositional logic and boolean algebra, De Morgan's laws are a pair of transformation rules that are both valid rules of inference. They are named after Augustus De Morgan, a 19th-century British mathematician. The rules allow the expression of conjunctions and disjunctions purely in terms of each other via negation.
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