Let $A$ and $B$ be two sets. For some set $C$, both $A\cap C$ and $B\cap C$ are empty sets and $A\cup C=B\cup C$. Which of the following is/are true ?
I. $C=\varphi$
II. $A=B$
III. $A\cup B=C$
Select the answer using the code given below :

Given $A \cup C = B \cup C$.

Taking intersection with $A$ on both sides:

$A \cap (A \cup C) = A \cap (B \cup C)$

$A = (A \cap B) \cup (A \cap C)$

Since $A \cap C = \emptyset$, we get $A = (A \cap B) \cup \emptyset = A \cap B$.

This implies $A \subseteq B$.

Similarly, taking intersection with $B$ on both sides of $A \cup C = B \cup C$:

$B \cap (A \cup C) = B \cap (B \cup C)$

$(B \cap A) \cup (B \cap C) = B$

Since $B \cap C = \emptyset$, we get $(A \cap B) \cup \emptyset = B \Rightarrow A \cap B = B$.

This implies $B \subseteq A$.

From $A \subseteq B$ and $B \subseteq A$, we conclude that $A = B$. Thus, statement II is true.

To check statements I and III, let $A = \{1\}$, $B = \{1\}$, and $C = \{2\}$.

Here, $A \cap C = \emptyset$ and $B \cap C = \emptyset$. Also, $A \cup C = \{1, 2\}$ and $B \cup C = \{1, 2\}$, so $A \cup C = B \cup C$.

However, $C \neq \emptyset$ and $A \cup B = \{1\} \neq C$.

Thus, statements I and III are not necessarily true. Only statement II is true.

Answer: II only