We are required to show that the following four conditions are equivalent:
\[ \text{(i) } A \subset B \] \[ \text{(ii) } A - B = \phi \] \[ \text{(iii) } A \cup B = B \] \[ \text{(iv) } A \cap B = A \]
To prove equivalence, it is sufficient to show that
\[ (i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (iv) \Rightarrow (i). \]
Proof that (i) ⇒ (ii):
Assume that \[ A \subset B. \]
Then every element of \( A \) is also an element of \( B \).
Hence, there is no element in \( A \) which is not in \( B \).
Therefore,
\[ A - B = \phi. \]
Proof that (ii) ⇒ (iii):
Assume that \[ A - B = \phi. \]
This means that there is no element of \( A \) outside \( B \), so every element of \( A \) belongs to \( B \).
Hence,
\[ A \subset B. \]
Therefore,
\[ A \cup B = B. \]
Proof that (iii) ⇒ (iv):
Assume that \[ A \cup B = B. \]
This implies that every element of \( A \) is already contained in \( B \), i.e.,
\[ A \subset B. \]
Therefore, the common elements of \( A \) and \( B \) are exactly the elements of \( A \).
Hence,
\[ A \cap B = A. \]
Proof that (iv) ⇒ (i):
Assume that \[ A \cap B = A. \]
This means that every element of \( A \) is also an element of \( B \).
Hence,
\[ A \subset B. \]
Since
\[ (i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (iv) \Rightarrow (i), \]
all the four conditions are equivalent.