Let us remind some representations whose represent in set theory:

A∪B={x∈A∨x∈B}

A∩B={x∈A∧x∈B}

A\B={x∈A,x∉B}

A^c={x∈U,x∉A}

A∆B=(A\B)∪(B\A)=(A∪B)\(A∩B)

Some important laws as follow:

(a)A∪B=B∪A (commutative law)

(b)A∪(B∪C)=(A∪B)∪C (associative law)

proof:

(a)

Let x be any element of the set A∪B,then

x∈(A∪B)

⇒x∈A∨x∈B

⇒x∈B∨x∈A

⇒x∈(B∪A)

Therefore,the conclusion is (A∪B)⊂(B∪A)

(b)

Let x be any element of the set A∪(B∪C),then

x∈A∪(B∪C)

⇔[x∈A]∨[x∈(B∪C)]

⇔[x∈A]∨[x∈B∨x∈C]

⇔[x∈A∨x∈B]∨[x∈C]

⇔x∈(A∪B)∪C

Therefore A∪(B∪C)=(A∪B)∪C

and we often express each side by writing simply A∪B∪C

Similarly,we can easy to prove A∩B=B∩A and A∩(B∩C)=(A∩B)∩C

The following laws are given(important):

commutative law:

(1)A∪B=B∪A

(2)A∩B=B∩A

associative law

(3)A∪(B∪C)=(A∪B)∪C

(4)A∩(B∩C)=(A∩B)∩C

distributive law of intersection over union:

(5)A∩(B∪C)=(A∩B)∪(A∩C)

distributive law ofunion over intersection:

(6)A∪(B∩C)=(A∪B)∩(A∪C)

De Morgan's law:

(7)A\(B∪C)=(A\B)∪(A\C)

(8)A\(B∩C)=(A\B)∩(A\C)

(9)(A∪B)^c=A^c∩B^c

(10)(A∩B)^c=A^c∪B^c

Try to prove!