#### Answer

A = {x ∈ Z | x = 6a + 4 for some integer a},
B = {y ∈ Z | y = 18b − 2 for some integer b}, and
C = {z ∈ Z | z = 18c + 16 for some integer c}
a. False, A ⊈ B.
Contradiction:
Suppose that x is a particular but arbitrarily chosen element of A and B.
By definition of A, there is an integer a such that x = 6a + 4.
By definition of B, there is an integer b such that x = 18b – 2.
By substitution, it follows that:
6a + 4 = 18b – 2
b = (6a + 4 + 2)/18 = (6a + 6)/18 = (a + 1)/3
Since b = (a + 1)/3, then b is not an integer since it is a quotient.
It follows that b is not an integer and b is an integer, which is a contradiction.
Therefore, the supposition is false and A ⊈ B.
Counterexample: There are elements in A that are not in B.
For example, if a = 3 is a particular but arbitrarily chosen element in A, then x = 6(3) + 4 = 22.
If x = 22 is an element in B, then:
22 = 18b – 2
b = (22 + 2)/18 = 24/18 = 6*4/(6*3) = 4/3
b = 4/3
Since b = 4/3, b is not an integer since it is a quotient.
It follows that b is not an integer and b is an integer, which is a contradiction.
Therefore, the supposition is false and A ⊈ B.
b. True, B⊆A
Proof: Suppose that y is a particular but arbitrarily chosen element of B.
By definition of B, there is an integer b such that y = 18b – 2.
[18b – 2 = 6a + 4
a = (18b – 6)/6 = 3b – 1]
Let a = 3b – 1.
Then a is an integer since it is a sum of integers.
By substitution, 6a + 4 = 6(3b – 1) + 4 = 18b – 6 + 4 = 18b – 2.
Therefore, by definition of A, x is an element of A.
c. True, B = C ↔ B⊆C ^ C⊆B
Proof that B⊆C:
Suppose that y is a particular but arbitrarily chosen element of B.
By definition of B, there is an integer b such that y = 18b – 2.
[18b – 2 = 18c + 16
c = (18b – 18)/18 = b – 1]
Let c = b - 1.
Then c is an integer since it is a sum of integers.
By substituting c into the definition of C, 18c + 16 = 18(b – 1) + 16 = 18b – 18 + 16 = 18b – 2 = x.
Therefore, by definition of C, x is an element of C.
Proof that C⊆B:
Suppose that z is a particular but arbitrarily chosen element of C.
By definition of C, there is an integer c such that z = 18c + 16.
[18c + 16 = 18b - 2
b = (18c + 18)/18 = c + 1]
Let b = c + 1.
Then b is an integer since it is a sum of integers.
By substituting b into the definition of B, 18b – 2 = 18(c + 1) – 2 = 18c + 18 – 2 = 18c + 16 = x.
Therefore, by definition of B, x is an element of B.