Unions and intersections: $(A \cup B = A ∪ C) \land (A \cap B = A ∩ C) \implies B = C.$ Prove or give a counterexample:
$$(A ∪ B = A ∪ C) \land (A ∩ B = A ∩ C) \implies B = C.$$
I think this is true, but I am not sure how to show it. I don't know if there are any manipulations with unions and intersections that allows me to move things around. Thanks in advance for any hints.
 A: We want to show that (i) every element of $B$ is an element of $C$ and (ii) every element of $C$ is an element of $B$. There is symmetry, so the proof of (ii) is almost the same as the proof of (i). We prove (i).
Proof of (i). Let $x\in B$. We want to show that $x\in C$.
Suppose first that $x$ is not an element of $A$. We have $x\in A\cup B$, so $x\in A\cup C$. But $x$ is not in $A$, so $x$ is in $C$.
Suppose next that $x$ is an  element of $A$. We have $x\in A\cap B$, therefore $x\in A\cap C$, and therefore $x\in C$.
A: Don't worry too much about tricks. If $x \in B$, then either $x \in B \cap A$ or $x \in B \setminus A$. Using those two cases, can you show that if $x \in B$, then $x \in C$ ?
If you can, then you can use that, and symmetry (ie swap $B$ for $C$), to establish the result.
A: $$\begin{align}
B &= B \cup (A \cap B)\\
&= B \cup (A \cap C)\\
&= (B \cup A) \cap (B \cup C)\\
&= (C \cup A) \cap (B \cup C)\\
&= (C \cup A) \cap (C \cup B)\\
&= C \cup (A \cap B)\\
&= C \cup (A \cap C)\\
&= C
\end{align}$$
A: To the earlier barrage of answers, here is an alternative in a different style: treat this as a simplification problem, and try to do this using the laws of logic, so on the level of elements/logic rather than on the set level, by starting out with expanding the definitions.
The number of symbols may seem a bit daunting at first, but the structure is really straightforward.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
This gives us:
$$\calc
    A \cup B = A \cup C \;\land\; A \cap B = A \cap C
\op\equiv\hint{set extensionality, twice}
    \langle \forall x :: x \in A \cup B \;\equiv\; x \in A \cup C \rangle
    \;\land\;
    \\&
    \langle \forall x :: x \in A \cap B \;\equiv\; x \in A \cap C \rangle
\op\equiv\hint{definitions of $\;\cup\;$ and $\;\cap\;$, each twice}
    \langle \forall x :: x \in A \lor x \in B \;\equiv\; x \in A \lor x \in C \rangle
    \;\land\;
    \\&
    \langle \forall x :: x \in A \land x \in B \;\equiv\; x \in A \land x \in C) \rangle
\op{\tag{*} \equiv}\hints{logic: LHS: extract common disjunct, see $\ref 0$ below;}
                   \hint{RHS: extract common conjunct, see $\ref 1$ below}
    \langle \forall x :: x \in A \;\lor\; (x \in B \equiv x \in C) \rangle
    \;\land\;
    \\&
    \langle \forall x :: x \in A \;\then\; (x \in B \equiv x \in C) \rangle
\op\equiv\hint{logic: write $\;P \then Q\;$ as $\;\lnot P \lor Q\;$; merge quantifications}
    \langle \forall x :: (x \in A \;\lor\; (x \in B \equiv x \in C))
    \;\land\;
    (x \not\in A \;\lor\; (x \in B \equiv x \in C)) \rangle
\op\equiv\hint{logic: extract common disjunct}
    \langle \forall x :: (x \in A \land x \not\in A) \;\lor\; (x \in B \equiv x \in C) \rangle
\op\equiv\hint{logic: left part is contradiction; simplify}
    \langle \forall x :: x \in B \equiv x \in C \rangle
\op\equiv\hint{set extensionality}
    B = C
\endcalc$$
The key step here of course is $\ref{*}$ which start the simplification phase of the calculation.  This step uses the following two laws of logic: $\;\lor\;$ distributes over $\;\equiv\;$, and its dual.
\begin{align}
\tag{0} P \lor (Q \equiv R) &\;\;\equiv\;\; (P \lor Q) \equiv (P \lor R)
\\
\tag{1} P \then (Q \equiv R) &\;\;\equiv\;\; (P \land Q) \equiv (P \land R)
\end{align}
A: Short answer: Draw Venn diagram of 3 sets.
Long answer: 
$$
B=(A\cap B\cap C)\cup(A\cap B\cap C^c)\cup(A^c\cap B\cap C)\cup(A^c\cap B\cap C^c)
$$
Observe that the second set in the union above is empty as can be seen here
$$
A\cap B\cap C^c = (A\cap B)\cap C^c = (A\cap C)\cap C^c = A\cap(C\cap C^c)=A\cap\emptyset=\emptyset.
$$.
Also observe that the last set in the union is also empty as
$$
A^c\cap B\cap C^c = B\cap(A^c\cap C^c)=B\cap(A\cup C)^c=B\cap(A\cup B)^c=B\cap A^c\cap B^c=\emptyset.
$$
Hence 
$$
B=(A\cap B\cap C)\cup(A^c\cap B\cap C)=B\cap C.
$$
By symmetry
$$
C=B\cap C
$$
Hence $B=C$.
A: Let $x\in B\setminus A$, then $x\in A\cup B$.  Hence, $x\in A\cup C$.  Since $x\notin A$, we must have $x\in C$. In other words, $B\setminus A\subseteq C\setminus A$.  A symmetric argument shows that $C\setminus A\subseteq B\setminus A$.
Now, 
$$B=(A\cap B)\cup(B\setminus A)=(A\cap C)\cup(C\setminus A)=C.$$
A: Any element in $A\cup B$ is either in $A$ or $B\setminus A$ exclusively (xor). Likewise any element in $A\cup C$ is either in $A$ or $C\setminus A$ exclusively.  When the unions are equal, then $B\setminus A=C\setminus A$.
$$A\cup B=A\cup C \vdash B\setminus A=C\setminus A$$
$B$ is the union of $B\setminus A$ and $B\cap A$.  We have shown that $B\setminus A$ is $C\setminus A$, and $B\cap A = C\cap A$ is given.  Then $B$ is the union of $C\setminus A$ and $C\cap A$; which is $C$.
Therefore $B=C$ when $A\cup B=A\cup C$ and $A\cap B=A\cap C$.
$$(A\cup B=A\cup C ) \wedge (A\cap B=A\cap C) \vdash (B=C)$$
A: 
$$B=((A\cup B) \backslash A)\cup (A\cap B)=((A\cup C)\backslash A)\cup (A\cap C) = C.$$
