We are required to show the following identities for any two sets \( A \) and \( B \):
\[ \text{(i) } A = (A \cap B) \cup (A - B) \]
\[ \text{(ii) } A \cup (B - A) = A \cup B. \]
Proof of (i):
Let \( x \in A \).
Then either \( x \in B \) or \( x \notin B \).
Case 1: If \( x \in B \), then
\[ x \in A \cap B. \]
Case 2: If \( x \notin B \), then
\[ x \in A - B. \]
Hence, in both cases,
\[ x \in (A \cap B) \cup (A - B). \]
Therefore,
\[ A \subseteq (A \cap B) \cup (A - B). \]
Conversely, let \[ x \in (A \cap B) \cup (A - B). \]
Then either \[ x \in A \cap B \quad \text{or} \quad x \in A - B. \]
In both cases, we have \[ x \in A. \]
Therefore,
\[ (A \cap B) \cup (A - B) \subseteq A. \]
Hence,
\[ A = (A \cap B) \cup (A - B). \]
Proof of (ii):
Consider the left-hand side:
\[ A \cup (B - A). \]
Let \( x \in A \cup (B - A) \).
Then either \[ x \in A \quad \text{or} \quad x \in B - A. \]
If \( x \in A \), then \[ x \in A \cup B. \]
If \( x \in B - A \), then
\[ x \in B \quad \text{and} \quad x \notin A, \]
which implies
\[ x \in A \cup B. \]
Hence,
\[ A \cup (B - A) \subseteq A \cup B. \]
Conversely, let \[ x \in A \cup B. \]
Then either \[ x \in A \quad \text{or} \quad x \in B. \]
If \( x \in A \), then \[ x \in A \cup (B - A). \]
If \( x \in B \) and \( x \notin A \), then
\[ x \in B - A, \]
which implies
\[ x \in A \cup (B - A). \]
Therefore,
\[ A \cup B \subseteq A \cup (B - A). \]
Hence,
\[ A \cup (B - A) = A \cup B. \]
Thus, both the identities are proved.