Question

I will mark as brainlist, It's urgent! plz answer as soon as possible!!

1.) State De Morgan’s laws and prove any one of them algebraically.
2.) From the premises x’=> y and y’=> z infer y’=>z

Answer 1

Answer:

(1)

Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)'

⇒ x ∉ (A U B)

⇒ x ∉ A and x ∉ B

⇒ x ∈ A' and x ∈ B'

⇒ x ∈ A' ∩ B'

⇒ x ∈ Q

Therefore, P ⊂ Q …………….. (i)

Again, let y be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A' ∩ B'

⇒ y ∈ A' and y ∈ B'

⇒ y ∉ A and y ∉ B

⇒ y ∉ (A U B)

⇒ y ∈ (A U B)'

⇒ y ∈ P

Therefore, Q ⊂ P …………….. (ii)

Now combine (i) and (ii) we get; P = Q i.e. (A U B)' = A' ∩ B'

Proof of De Morgan’s law: (A ∩ B)' = A' U B'

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

Therefore, 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

Therefore, N ⊂ M …………….. (ii)

Now combine (i) and (ii) we get; M = N i.e. (A ∩ B)' = A' U B'

Examples on De Morgan’s law:  

1. If U = {j, k, l, m, n}, X = {j, k, m} and Y = {k, m, n}.

Proof of De Morgan's law: (X ∩ Y)' = X' U Y'.

Solution:  

We know,  U = {j, k, l, m, n}

X = {j, k, m}

Y = {k, m, n}

(X ∩ Y) = {j, k, m} ∩ {k, m, n}            

          = {k, m}  

Therefore, (X ∩ Y)' = {j, l, n}  ……………….. (i)

Again, X = {j, k, m} so, X' = {l, n}

and    Y = {k, m, n} so, Y' = {j, l}

X' ∪ Y' = {l, n} ∪ {j, l}

Therefore,  X' ∪ Y' = {j, l, n}   ……………….. (ii)

Combining  (i)and (ii) we get;

(X ∩ Y)' = X' U Y'.          Proved

2. Let U = {1, 2, 3, 4, 5, 6, 7, 8}, P = {4, 5, 6} and Q = {5, 6, 8}.  

Show that (P ∪ Q)' = P' ∩ Q'.

Solution:

We know, U = {1, 2, 3, 4, 5, 6, 7, 8}

P = {4, 5, 6}

Q = {5, 6, 8}

P ∪ Q = {4, 5, 6} ∪ {5, 6, 8}  

        = {4, 5, 6, 8}

Therefore, (P ∪ Q)' = {1, 2, 3, 7}   ……………….. (i)

Now P = {4, 5, 6} so, P' = {1, 2, 3, 7, 8}

and Q = {5, 6, 8} so, Q' = {1, 2, 3, 4, 7}

P' ∩ Q' = {1, 2, 3, 7, 8} ∩ {1, 2, 3, 4, 7}

Therefore, P' ∩ Q' = {1, 2, 3, 7}   ……………….. (ii)

Combining  (i)and (ii) we get;

(P ∪ Q)' = P' ∩ Q'.          Proved

(2)

⇒ Q ≡ ¬ P∨ Q

¬(P∨Q) ≡ (¬P) ∧ (¬Q)

(P∧Q)∨(P∧R) ≡ P∧(Q∨R)

Explanation: