Proof 1

We will show that given a Set D, the Symmetric Difference Operator $\oplus$ over the set $P_{D} = {A : A \subseteq D}$ forms a group. So, we will show that: $< P_{D}, + >$ is a Group.

Definition: $A \oplus B = (A - B) \cup (B - A)$

Proof: We need to show that the Symmetric Difference Operator over $P_{D}$ : a) is Associative b) has an Identity Element c) has an Inverse Element

a) Associativity - In order to prove Associativity, we will show that $(A \oplus B) \oplus C = A \oplus (B \oplus C)$ Let us first show the LHS of the equation: $A \oplus B = (A - B) \cup (B - A)$ So, $(A \oplus B) \oplus C = ((A - B) \cup (B - A)) \oplus C$ Which is $(A \oplus B) \oplus C = (A - B - C) \cup (C - (A - B)$ Now, looking at the RHS of the equation: $B \oplus C = (B - C) \cup (C - B)$ So, $A \oplus (B \oplus C) = A \oplus ((B - C) \cup (C - B))$ Which is $A \oplus (B \oplus C) = (A - (B - C) \cup (C - B)) \cup ((B - C) \cup (C - B) - A)$ Now, this can be written, using $A \cup B = A + B - A \cap B$ : $A \oplus (B \oplus C) = (A - (B - C + C - B - (B - C) \cap (C - B))) \cup ((B - C + C - B - (B - C) \cap (C - B)) - A) = (A + (B - C) \cap (C - B)) \cup (- A - (B - C) \cap (C - B)) = (A - B - C) \cup (C - B + A) = (A \oplus B) \oplus C$ Therefore, the Symmetric Difference Operator is Associative.

2) Existence of Identity Element: For the identity element, we need the condition that: $A \oplus E = A$ So, by definition, $ϕ \subset P_{D}, \forall D$ Now, $A \oplus E = (A - E) \cup (A + E) = A$ We see that, if $E = ϕ$ , then, $A - E = A - ϕ = A$ Similarly, $E - A = ϕ - A = ϕ$ So, $A \oplus ϕ = (A - ϕ) \cup (ϕ - A) = A \cup ϕ = A$ Therefore, the Identity Element is $E = ϕ$

3) Inverse Element: For the Inverse Element $\forall A \in P_{D}, ∋, A \subset D$ , we need to show that $\exists A^{^{'}} \in P_{D}, ∋, A \oplus A^{^{'}} = E$ Now, $(A - A^{^{'}}) \cup (A^{^{'}} - A) = ϕ$ So, we need, $(A - A^{^{'}}) = ϕ$ and $(A^{^{'}} - A) = ϕ$ Therefore, this implies that $A^{^{'}} = A$ Therefore, the Inverse Element for an element A is the element itself.

Now, as $A \oplus B, \forall A, B \in P_{D}$ for some D is associative, has an identity element and an inverse element: $< P_{D}, \oplus >$ is a Group.

QED.