No, the statement "if $A \cup B \subset A \cup C$, then $B \subset C$" is not always true.

Counterexample:

Consider the following sets:

• $A = \{1\}$
• $B = \{2\}$
• $C = \{3\}$

Now, we check the union and subset relations:

• $A \cup B = \{1, 2\}$
• $A \cup C = \{1, 3\}$

Clearly, $A \cup B = \{1, 2\} \subset A \cup C = \{1, 3\}$ is false, and $B = \{2\} \not\subset C = \{3\}$, so the initial statement doesn't hold for these specific sets.

Explanation:

The condition $A \cup B \subset A \cup C$ only means that every element in $B$ is either in $C$ or in $A$. It does not necessarily imply that $B \subset C$. The element of $B$ might be present in $A$, making the inclusion $A \cup B \subset A \cup C$ true without requiring $B$ to be a subset of $C$.

Would you like further details or another example?

Here are some related questions:

1. What are the conditions for $A \cup B \subset A \cup C$ to hold?
2. Can you explain the relationship between $B$ and $C$ when $A \cup B \subset A \cup C$?
3. How can you prove a counterexample using different sets for $A$, $B$, and $C$?
4. What is the difference between the union and subset operations in set theory?
5. Can you apply the same logic to intersections of sets?

Tip: Always check set inclusions carefully by testing small examples to see if a general statement holds.