Set theory: Problem 3


Proof: Let \(x \in (A-B) \cup (B-C)\). Then by definition of union, \(x \in (A-B)\) or \(x \in (B-C)\). So we have the following cases:

So, in both cases, we have \(x \in A \cup B\) and \(x \notin A \cap B \cap C\). So \(x \in (A \cup B) - (A \cap B \cap C)\).

To summarize, we picked an arbitrary element x from \((A-B) \cup (B-C)\) and we have shown that it is also a member of \((A \cup B) - (A \cap B \cap C)\). By the definition of subset, this means that \((A-B) \cup (B-C) \subseteq (A \cup B) - (A \cap B \cap C)\).