THEORY OF AUTOMATA V2
ASSIGNMENT 1
11/9/2020
AHSAN HASSAN
F2018266093
Qno.1: PROVE EACH OF THE FOLLOWING
A) An(AuB)=A
PROOF ⊆: Claim An(AuB) ⊆ A
SUPPOSE x ∈ An(AuB)
So our supposition makes us believe that x∈A AND x∈(AuB)
The thing we are going to prove here is that x∈A
In particular x∈A so, this shows An(AuB) ⊆ A
Claim A⊆ An(AuB)
Suppose we have an element x that is in A so, x∈A then it also means that
x∈(AuB)
Also this means that x is eiher in A or in B
Then x∈A and x∈AuB , so x∈An(AuB)
This shows that A ⊆ An(AuB)
THUS BY HE METHOD OF DOUBLE INCLUSION WE COME TO KNOW THAT
An(AuB)=A(PROVED)
B) A-(BnC)=(A-B)u(A-C)
PROOF:
We know that any two sets P and Q are equal if and only if both are
subsets of each other, that is
P ⊂ Q and Q ⊂ P.
Let us consider that
x ∈ A - (B ∩ C)
⇒x∈A, x ∉ (B ∩ C)
⇒x∈A, (x∉B or x∉C)
⇒(x∈A, x∉B) or (x∈A, x∉C)
⇒x∈(A-B) or x∈(A-C)
⇒x∈(A-B) ∪ (A-C)
So, A - (B n C) ⊂ (A-B) U (A-C).
Again, let
x∈(A-B) ∪ (A-C)
⇒x∈(A-B) or x∈(A-C)
⇒(x∈A, x∉B) or (x∈A, x∉C)
⇒x∈A, (x∉B or x∉C)
⇒x∈A, x ∉ (B ∩ C)
⇒x ∈ A - (B ∩ C).
So, A - (B ∩ C) ⊂ (A-B) ∪ (A-C).
Therefore, we get
A - (B n C) = (A - B) U (A - C).
Hence proved.