Prove De Morgan’s law for (A∩ B)ʹ = Aʹ ∪ Bʹ
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Here is your answer:
To prove De Morgan's Law for ( A ∩ B )' = A' ∪ B'
Proof:
Let M = (A ∩ B)' and N = A' U B'
Let x be an arbitrary element of M
then x ∈ M ⇒ x ∈ (A ∩ B)'
⇒ x ∉ (A ∩ B)
⇒ x ∉ A or x ∉ B
⇒ x ∈ A' or x ∈ B'
⇒ x ∈ A' U B'
⇒ x ∈ N
⇔ M ⊂ N …………….. (i)
Again,
Let y be an arbitrary element of N
then y ∈ N ⇒ y ∈ A' U B'
⇒ y ∈ A' or y ∈ B'
⇒ y ∉ A or y ∉ B
⇒ y ∉ (A ∩ B)
⇒ y ∈ (A ∩ B)'
⇒ y ∈ M
⇔ N ⊂ M …………….. (ii)
Now,
From (i) and (ii) we get;
→ M = N
→ (A ∩ B)' = A' U B'
Hence proved
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