Prove the following: (i) (A \ B) \ C = (A \ C) \ B = A \ (B \cup C) (ii) A \cap (B \cup C) = (A \cap B) \cup (A \cap C) (iii) A \triangle (B \triangle C) = (A \triangle B) \triang... Prove the following: (i) (A \ B) \ C = (A \ C) \ B = A \ (B \cup C) (ii) A \cap (B \cup C) = (A \cap B) \cup (A \cap C) (iii) A \triangle (B \triangle C) = (A \triangle B) \triangle C Read more

Steps to Solve

1. Prove (i): (A \ B) \ C = (A \ C) \ B

To prove the equality of two sets, we show that each is a subset of the other.

First, we show that $(A \setminus B) \setminus C \subseteq (A \setminus C) \setminus B$. Let $x \in (A \setminus B) \setminus C$. This means $x \in (A \setminus B)$ and $x \notin C$. Since $x \in A \setminus B$, we have $x \in A$ and $x \notin B$. Thus, we know $x \in A$, $x \notin B$, and $x \notin C$. Since $x \in A$ and $x \notin C$, we have $x \in A \setminus C$. Since $x \in A \setminus C$ and $x \notin B$, we have $x \in (A \setminus C) \setminus B$. Therefore, $(A \setminus B) \setminus C \subseteq (A \setminus C) \setminus B$.

Next, we show that $(A \setminus C) \setminus B \subseteq (A \setminus B) \setminus C$. Let $x \in (A \setminus C) \setminus B$. This means $x \in (A \setminus C)$ and $x \notin B$. Since $x \in A \setminus C$, we have $x \in A$ and $x \notin C$. Thus, we know $x \in A$, $x \notin B$, and $x \notin C$. Since $x \in A$ and $x \notin B$, we have $x \in A \setminus B$. Since $x \in A \setminus B$ and $x \notin C$, we have $x \in (A \setminus B) \setminus C$. Therefore, $(A \setminus C) \setminus B \subseteq (A \setminus B) \setminus C$. Since both inclusions hold, we conclude that $(A \setminus B) \setminus C = (A \setminus C) \setminus B$.

2. Prove (i): (A \ B) \ C = A \ (B ∪ C)

Now, we show that $(A \setminus B) \setminus C = A \setminus (B \cup C)$.

First, we show that $(A \setminus B) \setminus C \subseteq A \setminus (B \cup C)$. Let $x \in (A \setminus B) \setminus C$. This means $x \in (A \setminus B)$ and $x \notin C$. Since $x \in A \setminus B$, we have $x \in A$ and $x \notin B$. Thus, we know $x \in A$, $x \notin B$, and $x \notin C$. Since $x \notin B$ and $x \notin C$, we have $x \notin (B \cup C)$. Since $x \in A$ and $x \notin (B \cup C)$, we have $x \in A \setminus (B \cup C)$. Therefore, $(A \setminus B) \setminus C \subseteq A \setminus (B \cup C)$.

Next, we show that $A \setminus (B \cup C) \subseteq (A \setminus B) \setminus C$. Let $x \in A \setminus (B \cup C)$. This means $x \in A$ and $x \notin (B \cup C)$. Since $x \notin (B \cup C)$, we have $x \notin B$ and $x \notin C$. Thus, we know $x \in A$, $x \notin B$, and $x \notin C$. Since $x \in A$ and $x \notin B$, we have $x \in A \setminus B$. Since $x \in A \setminus B$ and $x \notin C$, we have $x \in (A \setminus B) \setminus C$. Therefore, $A \setminus (B \cup C) \subseteq (A \setminus B) \setminus C$. Since both inclusions hold, we conclude that $(A \setminus B) \setminus C = A \setminus (B \cup C)$.

Combining the two results, we have $(A \setminus B) \setminus C = (A \setminus C) \setminus B = A \setminus (B \cup C)$.

3. Prove (ii): A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

This is the distributive property of intersection over union. First, we show that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$. Let $x \in A \cap (B \cup C)$. This means $x \in A$ and $x \in (B \cup C)$. Since $x \in B \cup C$, either $x \in B$ or $x \in C$ (or both). If $x \in B$, then since $x \in A$, we have $x \in A \cap B$. If $x \in C$, then since $x \in A$, we have $x \in A \cap C$. Therefore, $x \in (A \cap B) \cup (A \cap C)$. Thus, $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$.

Next, we show that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$. Let $x \in (A \cap B) \cup (A \cap C)$. This means $x \in (A \cap B)$ or $x \in (A \cap C)$ (or both). If $x \in A \cap B$, then $x \in A$ and $x \in B$. Since $x \in B$, we have $x \in B \cup C$. Thus $x \in A \cap (B \cup C)$. If $x \in A \cap C$, then $x \in A$ and $x \in C$. Since $x \in C$, we have $x \in B \cup C$. Thus $x \in A \cap (B \cup C)$. Therefore, $x \in A \cap (B \cup C)$. Thus, $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.

Since both inclusions hold, we conclude that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

4. Prove (iii): A Δ (B Δ C) = (A Δ B) Δ C

To prove this, we recall that $A \triangle B = (A \setminus B) \cup (B \setminus A)$. Also $A \triangle B = (A \cup B) \setminus (A \cap B)$. We have $A \triangle (B \triangle C) = A \triangle ((B \setminus C) \cup (C \setminus B)) = (A \setminus ((B \setminus C) \cup (C \setminus B))) \cup (((B \setminus C) \cup (C \setminus B)) \setminus A)$. Similarly, $(A \triangle B) \triangle C = ((A \setminus B) \cup (B \setminus A)) \triangle C = (((A \setminus B) \cup (B \setminus A)) \setminus C) \cup (C \setminus ((A \setminus B) \cup (B \setminus A)))$.

Alternatively, we can use the property $A \triangle B = (A \cup B) \setminus (A \cap B)$. Using this, we have $A \triangle (B \triangle C) = (A \cup (B \triangle C)) \setminus (A \cap (B \triangle C))$. And $(A \triangle B) \triangle C = ((A \triangle B) \cup C) \setminus ((A \triangle B) \cap C)$.

Another approach is to define the characteristic function $\chi_A(x)$ to be $1$ if $x \in A$ and $0$ if $x \notin A$. Then $\chi_{A \triangle B}(x) = \chi_A(x) + \chi_B(x) \pmod{2}$. Therefore, $\chi_{A \triangle (B \triangle C)}(x) = \chi_A(x) + \chi_{B \triangle C}(x) \pmod{2} = \chi_A(x) + \chi_B(x) + \chi_C(x) \pmod{2}$. Also, $\chi_{(A \triangle B) \triangle C}(x) = \chi_{A \triangle B}(x) + \chi_C(x) \pmod{2} = \chi_A(x) + \chi_B(x) + \chi_C(x) \pmod{2}$. Since the characteristic functions are equal, the sets are equal, so $A \triangle (B \triangle C) = (A \triangle B) \triangle C$.