A b by definition of set difference in either case x

∈ A - B by definition of set difference. In either case, x / ∈ A - B
and so x ∈ A - B . Therefore, A ∪ B ⊂ A - B .
Putting both statements A - B ⊂ A ∪ B and A ∪ B ⊂ A - B together we have proved A - B =
A ∪ B .
ii. Using Laws of Sets:
( A ∩ B ) ∪ ( A ∩ B ) = A ∩ ( B ∪ B ) Distributive Law
= A ∩ U Complement Law
= A
Using Subset Containment:
Proof. We first prove ( A ∩ B ) ∪ ( A ∩ B ) ⊂ A . Let x be a specific but arbitrarily-chosen element of
( A ∩ B ) ∪ ( A ∩ B ) . Then x ∈ A ∩ B or x ∈ A ∩ B . Consider two cases:
1. x ∈ A ∩ B : By definition we have x ∈ A .
2. x ∈ A ∩ B : By definition we have x ∈ A .

Homework 3: Introductory Set Theory Solutions
(c) Proof. Suppose by way of contradiction that B ⊂ A and A 6⊂ B . Therefore there exists x ∈ A such that
x / ∈ B . However this means x ∈ B . By hypothesis, B is a subset of A so x ∈ A as well. But then x / ∈ A
and x ∈ A is a contradiction.
(d) The statement “ It follows that x / ∈ A or x / ∈ B by definition of complement, and so x / ∈ A ∪ B by definition
of union ” is logically incorrect. Imagine the statement