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

Explanation:

The statement says that if the union of set $A$ with set $B$ is a subset of the union of set $A$ with set $C$, then set $B$ must be a subset of set $C$.

However, this reasoning is flawed because the set $A$ could overlap with $B$ in such a way that elements in $B$ that are also in $A$ don't need to be in $C$ for $A \cup B \subseteq A \cup C$ to hold. To show that this is false, let's look at a counterexample:

Counterexample:

Let:

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

Then:

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

Clearly, $A \cup B = \{1, 2\} \subseteq \{1, 3\} = A \cup C$, so the premise $A \cup B \subseteq A \cup C$ is true.

But $B = \{1, 2\}$ is not a subset of $C = \{3\}$, since $2 \in B$ and $2 \notin C$.

Thus, the implication $B \subseteq C$ does not hold, proving that the original statement is false.

Would you like further clarification on any part of this reasoning?

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Here are 5 related questions to deepen your understanding:

1. What are the properties of set unions and subsets?
2. How can counterexamples be used to disprove logical statements?
3. What is the difference between strict subset and non-strict subset?
4. Can the inclusion $A \cup B \subseteq A \cup C$ imply something about the relationship between $B$ and $C$ in specific cases?
5. How do intersections of sets relate to unions in this context?

Tip: When trying to prove or disprove set relationships, constructing counterexamples is a powerful tool for testing the limits of the statement.