Prove that if $A – C = B – C$ and $A\cap C = B\cap C$ then $A = B$ I am trying to prove that if $A – C = B – C$ and $A\cap C = B\cap C$ then $A = B$. I have tried using Venn Diagrams as a proof technique, but we are not able to use proof by Venn Diagrams.
 A: By symmetry of the question, it suffices to show $A\subset B$. (The proof will work for $B\subset A$.) Let $x\in A$; we need to show $x\in B$.
Step 1: Suppose $x\in C$. Then $x$ is in the intersection $A\cap C$. Conclude with the hypothesis.
Step 2: Suppose $x\notin C$. Then $x\in A-C$. Again we can conclude easily.
A: Let $x\in A$. We have two cases:


*

*$x\in C$. Then $x\in A\cap C=B\cap C$, hence $x\in B$. 

*$x\notin C$. Then $x\in A-C=B-C$, hence $x\in B$


Same goes for the opposite direction. 
A: $A= (A \cap C) \cup (A \cap C^C) = (A \cap C) \cup (A-C) = (B \cap C) \cup (B -C) = (B \cap C) \cup (B \cap C^C)=B$
A: In a set theory proof of $A=B$, you must prove two statements:


*

*If $x\in A$, then $x\in B$.  (This proves $A\subseteq B$.)

*If $x\in B$, then $x\in A$.  (This proves $B\subseteq A$.)
Let's try to prove that if $x\in A$, then $x\in B$.  So, we assume that $x\in A$, and our goal is to conclude that $x\in B$.  Now, there are two cases, either $x\in C$ or $x\not\in C$.  We can consider each case separately.
If $x\in C$, then since $x\in A$ was assumed, $x\in A\cap C$.  Since $A\cap C=B\cap C$, then $x\in B\cap C$.  Hence $x\in B$ and $x\in C$.  Therefore, $x\in B$.
If $x\not\in C$, then since $x\in A$ was assumed, $x\in A-C$, but since $A-C=B-C$, $x\in B-C$.  Therefore, $x\in B$ and $x\not\in C$.  Therefore, $x\in B$.
Since either case leads to $x\in B$, we know that $x\in A$ implies that $x\in B$.  Hence $A\subseteq B$.
Now, you must prove the other direction.
A: Let $x \in X$.  
If $x \in A$ and $x \in C$ then $x \in A\cap C = B \cap C$ so $x \in A$.
if $x \in A$ and $x \not \in C$ then $x \in A \setminus C = B \setminus C$ so $x \in B$.
if $x \not \in A$ and $x \in C$ then $x \not \in A\cap C=B\cap C$. So either $x \not \in B$ or $x \not \in C$.  But $x \in C$ so $x \not \in B$.
If $x \not \in A$ and $x \not \in C$ then $x \not \in A \setminus C$.  So if $x \in B$ then $x \not \in C$ so $x \in B \setminus C = A \setminus C$.  That's not possible so $x \not \in B$.
So $x \in A \iff x \in B$.  So $A = B$
.... OR ..... $(D \setminus E)\cup (D \cap E) = \{x|x \in D; x \not \in E\} \cup \{x|x \in D; x \in E\} = \{x|x \in D; $ and either $x \in E$ or $x \not \in E\}=$ .... but all $x$ are either in $E$ or not in $E$ ... $= \{x|x \in D \} = D$.
Therefore $A = (A\setminus C) \cup (A\cap C) = (B\setminus C) \cup (B\cap C) = B$.
......
or
If $x \in A$ then if $x \in C$ $x \in A\cap C = B\cap C$ so $x \in B$.
If $x \not \in C$ then $x \in A \setminus C = B\setminus C$ so $x \in B$.
So $x \in A\implies x \in B$ so $A \subset B$.
Do the same to show $B \subset A$.  So $A \subset B$ and $B \subset A$.  So $A = B$.
