# Set theory: Problem 3

### Solution

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:

• Case 1: $$x \in (A-B)$$. Then by definition of set difference, $$x \in A$$ and $$x \not\in B$$. Since $$A \subseteq A \cup B$$, $$x \in A$$ implies that $$x \in A \cup B$$. Since $$A \cap B \cap C \subseteq B$$, $$x \not\in B)$$ implies that $$x \notin A \cap B \cap C$$.
• Case 2: $$x \in (B-C)$$. Then $$x \in B$$ and $$x \not\in C$$. Since $$B \subseteq A \cup B$$, $$x \in B$$ implies that $$x \in A \cup B$$. Since $$A \cap B \cap C \subseteq C$$, $$x \not\in C$$ implies $$x \notin A \cap B \cap C$$.

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 $$(A-B) \cup (B-C) \subseteq (A \cup B) - (A \cap B \cap C)$$.