Set Theory Rules are as follows

(A∪A) = A,  (A∩A) = A (idempotent law)
(A∪B)’ = A’∩B’, (A ∩ B) ′ = A ′ ∪ B ′ (De Morgan’s law)
A∪ B = B ∪ A, A∩B = B∩A (commutative law)
(A ∩ B) ∩ C = A ∩ (B ∩ C), (A ∪ B) ∪ C = A ∪ (B ∪ C) (associative law)
A ∪ φ = A, A ∩ U = A, A ∪ U = U, A ∩ φ = φ (identity law)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C), A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)   (distributive law)
(A’)’ = A (involution law)

Basic Set Theory Properties are:
A st is inherently unordered. This means that the order of elements in a st does not make a new st. Change in order of elements inside a st is feasible. For example if st A = {a, x, c} then A = {a, c, x} = {x, c, a} = {x, a, c} = {c, a, x} = {c, x, a}.
Each element in a st is distinct. Multiple repetition of an element makes no difference. For example if st A = {2, 3} then A = {2, 2, 3, 3} = {2, 3, 3} = {2, 2, 3} = {2, 2, 2, 3, 3, 3, 3} = {2, 3} and so on.
Above rules and properties of Set Theory Help you to solve Set Theory Questions easily.
Let us see some Set Theory Problems and Solutions now:

Q.1) prove that (A ′ ∩ B) ′ ∩ (A ∪ B) = A
Sol.) lets begin with LHS of the above problem:
LHS = ((A ′) ′ ∪ B ′) ∩ (A ∪ B) (by De Morgan’s law)
= (A∪ B’) ∩ (A ∪ B) (by involution law)
= A∪ (B∩ B’) (distributive law/property)
= A ∪ φ (complement law)
= A (identity law) = RHS
Hence, proved.

Q.2) in a community if 70 % people can speak English and 60% people can speak French then find the number of people who can speak both languages.
Sol.) Number of people who can speak English = n(E) = 70 %
Number of people who can speak French = n(F) = 60%
Number of people who can speak both languages will be n(E∩F) = ?
Total number of people in the community n (E∪F)= 100%
Now n (E∪F)= n(E) + n(F) - n(E∩F)
100=70 + 60 - n(E∩F)
100=130 - n(E∩F)
n(E∩F)=30%