$A, B,$ and $C$ are all sets that lie in a common universal set. Prove $A = B$ given the following statements. $(A \cup C) = (B \cup C)$ and $(A \cap C) = (B \cap C)$
I know I have to start with $A = ...$ to get to $B$. I incorporated $A = A \cap (A \cup C)$ as the first step using absorption law, but I don't know where to go from here. How do I show that it equals $(A \cup C) = (B \cup C)$? I've looked at the set identities, and think that I have to use absorption law again to show that $A \cap (A \cup C) = (A \cap (A \cup C)) \cap ((A \cap (A \cup C)) \cup (B \cup C))$. Can you guys please help me solve this problem, I'm stuck.
 A: If the element is in $A$ then either it's in $C$ or it isn't.
If it is in $C$ then it is in their intersection, which implies that it is in the intersection of $B$ and $C$ which implies that it is in $B$.
If it is not in $C$, it is still in the union of $A$ and $C$, which implies that it is in the union of $B$ and $C$, which implies that it is in $B$ (since it is assumed here that it is not in $C$).
By a symmetrical argument, $B$ is also a subset of $A$.
A: You can use absorption and distributivity:
\begin{align}
A&=A\cap(A\cup C)&&\text{absorption}\\
&=A\cap(B\cup C)&&\text{assumption}\\
&=(A\cap B)\cup(A\cap C)&&\text{distributivity}\\
&=(A\cap B)\cup(B\cap C)&&\text{assumption}\\
&=(B\cap A)\cup(B\cap C)&&\text{commutativity}\\
&=B\cap(A\cup C)&&\text{distributivity}\\
&=B\cap(B\cup C)&&\text{assumption}\\
&=B&&\text{absorption}
\end{align}
You can see that this holds in every distributive lattice.
A: Assuming $(A \cup C) = (B \cup C)$ and $(A \cap C) = (B \cap C)$. Prove $A=B$.
Proof by Contradiction:
Assume $x \in A$ and $x \notin B$.
Then $x \in B \cup C \to x \in$ C.
But this leads to a contradiction. Because $x \notin B \cap C$ but $x \in  A \cap C$.
Therefore, if $x \in A$ then $x \in B$.
And the same argument works for $x \in B$ but $x \notin A$.
Therefore, if something is in $A$ then it is in $B$ and vice versa.
A: Let me try a direct approach. For sets $A$ and $C$, we have one of the following two possibilities:

*

*$A \subset C$, this includes the case where both sets are equal;


*$A \cap C$ is some proper subset of $A$, which may be empty.
For the first case, we know that $A \cup C = C = B \cup C$, hence $B \subset C$. By the second equality given in the question, we have $A = A \cap C = B \cap C = B$.
For the second case, let $A \cap C = D = B \cap C$. By the first equality in the question, $(A \cup C) \backslash C = (B \cup C) \backslash C \implies A \backslash D = B \backslash D$. Taking union with $D$ on both sides gives us the desired result.
A: I suggest a Venn diagram of $A, B, C$.
First solution.
Let $S_1=A$ \ $(B\cup C)$ and $S_2=B$ \ $(A\cup C)$ and $S_3=C$ \ $(A\cup B).$
Let $S_4=(A\cap B)$ \ $C$ and $S_5=(A\cap C)$ \ $B$ and $S_6=(B\cap C)$ \ $A .$
Let $S_7=A\cap B\cap C.$
Then $\{S_j:1\le j\le 7\}$ is a pairwise-disjoint family.
For brevity let $T=\cup_{j=3}^{j=7}\,S_j.$ Then $S_1\cup T=A\cup C=B\cup C=S_2\cup T.$ Therefore $$S_1=S_1\cap (S_1\cup T)=S_1\cap (S_2\cup T)=\emptyset. $$ Interchanging the subscripts $1, 2$ in the preceding line, we also have $S_2=\emptyset.$
We have $$S_5=S_5\cap ( S_5\cup S_7)=S_5\cap (A\cap C)=S_5\cap (B\cap C)=S_5\cap (S_6\cup S_7)=\emptyset.$$ And similarly we obtain $S_6=\emptyset.$
Therefore $$A=S_1\cup S_5\cup S_4\cup S_7=S_2\cup S_6\cup  S_4\cup S_7 =B.$$
Second solution.
For any $A,B,C$ we have $$A=(A\cap C)\cup (A\setminus C)=(A\cap C)\cup ([A\cup C]\setminus C)$$ and $$B=(B\cap C)\cup (B\setminus C)=(B\cap C)\cup ([B\cup C]\setminus C)$$. Now compare the RHS of these if $A\cap C =B\cap C$ and $A\cup C=B\cup C.$
