# Compliments Proof 4

Let $X \subset Y \subset Z$. Prove that $Z - (Y - X) = X \cup (Z - Y)$.

We will begin by letting $A = Z - (Y - X)$ and letting $B = X \cup (Z - Y)$. If $x \in A \Rightarrow x \in Z, x \notin Y - X$, so $Z - (Y - X) = Z \cap C(Y - X)$, and $Z \cap C(Y - X) = Z \cap (C(Y) \cap C(X))$ and : $Z \cap C(Y \cap C(X)) = Z \cap (C(Y) \cup X) = (Z \cap C(Y)) \cup (Z \cap X)$ since $Z \cap C(Y) = Z - Y$ and $Z \cap X = X$, it follows that $Z \cap C(Y - X) = Z \cap C(Y \cap C(X))$

This came from http://ashleymills.com/.

# Compliments Proof 3

Let $X \subset Y \subset Z$. Prove that $C_{Y}(X) \subset C_{Z}(X)$.

In order for $C_{Y}(X) \subset C_{Z}(X), \forall_{x \in C_{Y}(X)}(x \in C_{Z}(X)) \Rightarrow \exists_{x \notin X} : x \in Y \cup Z$ because $X \subset Y \subset Z$. If we assume that $\forall_{x \in X}(x \in Y \cup Z)$ then $C_{Y}(X)$ and $C_{Z}(X)$ would both be $\emptyset$ and $\emptyset \not\subset \emptyset$. $\therefore \exists_{x \notin X} : x \in Y \cup Z$. Also $\not\exists_{x \in X}: x \notin Y \cup Z$ because then $X \not\subset Y$ which creates a contradiction with the axiom above. Since $\exists_{x \in Y \cup Z}: x \notin X$, $\forall_{x \in C_{Y}(X)}(x \in C_{Z}(X)) \Rightarrow C_{Y}(X) \subset C_{Z}(X)$.

# Compliments Proof 2

Let $A \subset S$ and $B \subset S$. Prove that $A \subset C(B)$ if and only if $B \subset C(A)$.

Let’s begin by letting $A \subset C(B)$ and assume that $B \not\subset C(A)$. From this assumption it follows that $\exists_{x \in B} : x \notin C(A) \Rightarrow \exists_{x \notin C(A)} : x \in B$. This means that $\exists_{x \in A} : x \notin C(B) \Rightarrow A \not\subset C(B)$. This creates a contradiction with the above statement and therefore $B \subset C(A)$. This proves the second half of the statement, and now we must prove the first half of the statement.

We will now let $B \subset C(A)$ and assume that $A \not\subset C(B)$. From this assumption it follows that $\exists_{x \in A} : x \notin C(B) \Rightarrow \exists_{x \notin C(B)} : x \in A$. This means that $\exists_{x \in B} : x \notin C(A) \Rightarrow B \not\subset C(A)$ creating a contradiction with the above statement and therefore $A \subset C(B)$.

# Compliments Proof

Let $A \subset S$, $B \subset S$. Prove that $A \subset B$ if and only if $C(B) \subset C(A)$

Let $C(B) \subset C(A)$ and assume that $A \not\subset B$. From this assumption it follows that $\exists_{x\in A} : x\notin B$, which means that $\exists_{x\notin B} : x\in A$, which can also be stated as $\exists_{x\in C(B)} : x \notin C(A) \therefore C(B) \not\subset C(A)$ which contradicts with the above assumption and therefore $A \subset B$. Now if we let $A \subset B$ and assume that $C(A) \not\subset C(B)$ than it follows that $\exists_{x\in C(A)} : x \notin C(B) \Rightarrow \exists _{x\in A} : x \notin B$. This means that $A \notin B$ which contradicts the above statement and therefore $C(A) \subset C(B)$. This came from  http://ashleymills.com/.