Understanding Set Equality Using Symmetric Difference and Intersection

This problem asks us to determine the relationship between sets B and C, given two specific conditions involving a third set A. The conditions are:

• Condition 1: The symmetric difference of sets A and B is equal to the symmetric difference of sets A and C. This is written mathematically as $A \Delta B = A \Delta C$.
• Condition 2: The intersection of sets A and B is equal to the intersection of sets A and C. This is written as $A \cap B = A \cap C$.

We need to find the conclusion that logically follows from these two statements about set equality.

Method 1: Leveraging Symmetric Difference Properties

The symmetric difference operation ($\Delta$) is fundamental here. Recall that $X \Delta Y$ contains elements that are in either set X or set Y, but not in both. This operation has several key properties that are useful for simplifying equations:

• Associativity: $ (X \Delta Y) \Delta Z = X \Delta (Y \Delta Z) $ for any sets X, Y, Z.
• Identity Element: The empty set ($\emptyset$) acts as the identity, meaning $ X \Delta \emptyset = X $.
• Inverse Property: Every set is its own inverse, meaning $ X \Delta X = \emptyset $.

Let's start with the first condition given: $A \Delta B = A \Delta C$. We can use the properties of the symmetric difference to simplify this equation. A common technique is to take the symmetric difference of both sides with set A:

$ A \Delta (A \Delta B) = A \Delta (A \Delta C) $

Now, we apply the associative property to rearrange the terms on each side:

$ (A \Delta A) \Delta B = (A \Delta A) \Delta C $

Using the inverse property, we know that $A \Delta A$ equals the empty set, $\emptyset$:

$ \emptyset \Delta B = \emptyset \Delta C $

Finally, applying the identity property ($\emptyset \Delta X = X$), we get:

$ B = C $

This method shows that the first condition, $A \Delta B = A \Delta C$, is powerful enough on its own to conclude that $B = C$. The second condition ($A \cap B = A \cap C$) is actually implied by the first one, because if $B$ and $C$ are identical, their intersections with $A$ must also be identical.

Method 2: Detailed Proof Using Set Definitions

To provide a more detailed explanation, let's use the definition of symmetric difference: $X \Delta Y = (X \setminus Y) \cup (Y \setminus X)$, or alternatively, $X \Delta Y = (X \cap Y^c) \cup (X^c \cap Y)$. We will use the latter form and consider both given conditions.

Given:

1. $A \Delta B = A \Delta C \implies (A \cap B^c) \cup (A^c \cap B) = (A \cap C^c) \cup (A^c \cap C)$
2. $A \cap B = A \cap C$

Our goal is to show $B = C$. This involves proving two subset relations: $B \subseteq C$ and $C \subseteq B$.

Proof of $B \subseteq C$:

Let $x$ be an arbitrary element. Assume $x \in B$. We need to show that $x$ must also be in $C$. We consider two cases:

• Case 1: $x \in A$. If $x \in A$ and we assumed $x \in B$, then $x$ belongs to the intersection $A \cap B$. From Condition 2 ($A \cap B = A \cap C$), it follows that $x$ must also be in $A \cap C$. If $x \in A \cap C$, then $x \in C$.
• Case 2: $x \notin A$. If $x \notin A$ and we assumed $x \in B$, then $x$ belongs to $A^c \cap B$. According to the definition of symmetric difference, $A^c \cap B$ is a part of $A \Delta B$. Since Condition 1 states $A \Delta B = A \Delta C$, $x$ must also be an element of $A \Delta C$. The definition $A \Delta C = (A \cap C^c) \cup (A^c \cap C)$ means $x$ must be in either $A \cap C^c$ or $A^c \cap C$. Because we are in the case where $x \notin A$, $x$ cannot be in $A \cap C^c$. Therefore, $x$ must be in $A^c \cap C$. If $x \in A^c \cap C$, it directly implies $x \in C$.

Since in both cases ($x \in A$ or $x \notin A$), we found that $x \in C$, we have successfully shown that if $x \in B$, then $x \in C$. Thus, $B \subseteq C$.

Proof of $C \subseteq B$:

Similarly, let $y$ be an arbitrary element. Assume $y \in C$. We need to show that $y$ must also be in $B$. We again consider two cases:

• Case 1: $y \in A$. If $y \in A$ and we assumed $y \in C$, then $y$ belongs to the intersection $A \cap C$. From Condition 2 ($A \cap C = A \cap B$), it follows that $y$ must also be in $A \cap B$. If $y \in A \cap B$, then $y \in B$.
• Case 2: $y \notin A$. If $y \notin A$ and we assumed $y \in C$, then $y$ belongs to $A^c \cap C$. According to the definition of symmetric difference, $A^c \cap C$ is a part of $A \Delta C$. Since Condition 1 states $A \Delta C = A \Delta B$, $y$ must also be an element of $A \Delta B$. The definition $A \Delta B = (A \cap B^c) \cup (A^c \cap B)$ means $y$ must be in either $A \cap B^c$ or $A^c \cap B$. Because we are in the case where $y \notin A$, $y$ cannot be in $A \cap B^c$. Therefore, $y$ must be in $A^c \cap B$. If $y \in A^c \cap B$, it directly implies $y \in B$.

Since in both cases ($y \in A$ or $y \notin A$), we found that $y \in B$, we have successfully shown that if $y \in C$, then $y \in B$. Thus, $C \subseteq B$.

Having proven both $B \subseteq C$ and $C \subseteq B$, we can definitively conclude that $B = C$. This detailed analysis confirms the result obtained more quickly using the algebraic properties of the symmetric difference.

Conclusion

Given the conditions $A \Delta B = A \Delta C$ and $A \cap B = A \cap C$, the relationship that logically follows is $B = C$. Both methods of proof confirm this result.