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))

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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/.