Help with Set Theory/ Proofs Can you conclude that $A = B$ if $A$, $B$, and $C$ are sets such that
(a) $A \cup C = B \cup C$
No, the sets $A=\{1,2\}, B=\{3,4\}, C=\{1,2,3,4,5\}$ disprove this, because $A \cup C = B \cup C$ but $A\neq B$
(b) $A \cap C = B \cap C$
No, the sets $A=\{1,2\}, B=\{1,4\}, C=\{1\}$ disprove this because $A \cap  C = B \cap C=\{1\}$, however $A\neq B$
(c) $A \cup C = B \cup C$ and $A \cap C = B \cap C$
Having Trouble with (c)
This is what I have based on your response
L1 A ∪ C = B ∪ C
L2 A ∩ C = B ∩ C
L3 ∀x(x ∈ A ∪ x ∈ C = x ∈ B ∪ x ∈ C    )    
L4 ∀x(x ∈ A ∪ x ∈ C = x ∈ B ∪ x ∈ C    )    
L5 ∀x(x ∈ A ∩ x ∈ C = x ∈ B ∩ x ∈ C)
L6 ∀x(x ∈ A ∩ x ∉ C = x ∈ B ∩ x ∉ C)

 A: We are given sets $A,B$ and $C$ such that $A \cup C = B \cup C$ and $A \cap C = B \cap C$. We wish to conclude $A = B$.
To show $A = B$ it suffices to show $A \subseteq B$ and $B \subseteq A$. 
I will go through one of these arguments leaving you to show the other.
So, let's show that $A \subseteq B$, i.e. for every $a$ in $A$ we have $a$ in $B$:
Since $a \in A$ and $A \subseteq A \cup C$ it follows that $a \in A \cup C$. Now since $A \cup C = B \cup C$ if follows that $a \in B \cup C$. 
Ok, now we ask a question, is $a$ in $C$? Well, it might be or it might not be. But either it is in $C$ or it is not. So there are two cases:
1.) Suppose $a$ is in $C$. Then $a$ is in $A$ and $C$, i.e. $a \in A \cap C$. Since $A \cap C = B \cap C$ (by our assumption) it follows that $a \in B \cap C$, in particular $a$ is in $B$.
2.) Suppose $a$ is not in $C$. Ok, well, we know that $a$ is in $B \cup C$, so if $a$ is not in $C$ then it must be in $B$. 
Thus, in both cases $a$ is in $B$, which is exactly what we wanted. Thus we have shown $A \subseteq B$.
Now you must use a similar argument to show that $B \subseteq A$. When you have this, it follows (by definition of set-equality) that $A = B$.
A: Let it be that $A\cup C=B\cup C$ and $A\cap C=B\cap C$ and assume
that $a\in A\wedge a\notin B$.
Then $a\in A\cup C=B\cup C$ and $a\notin B$ tells us that $a\in C$. 
However, then $a\in A\cap C=B\cap C\subset B$ and a contradiction
is found. 
We conclude that $\forall x\;\left[x\notin A\vee x\in B\right]$ or
equivalently $A\subseteq B$. 
Likewise we find $B\subseteq A$ hence $A=B$.
A: We know $A = B$ iff both the following:
$$x \in A \implies x \in B\\
x \in B \implies x \in A
$$
We also know
$$
x\in A \implies x \in A\cup C
$$
$$
x \in A\cap C \implies x \in A, x\in C
$$
$$
\begin{align}
x\in A \cup C \implies &x \in A, \text{or}\\
&x\in C
\end{align}
$$
which transposes to:
$$
x\not\in C \implies (x\in A\cup C \implies x \in A)
$$
We assume:
$$
A\cap C = B \cap C,\\
A\cup C = B\cup C
$$
Select $x \in A$.  Then either $x\in C$ or $x\not\in C$.


*

*$x\in C$:
$$
\begin{align}
x\in A,x\in C &\implies x\in A\cap C\\
&\implies x\in B\cap C\\
&\implies x\in B
\end{align}
$$

*$x\not\in C$
$$
\begin{align}
x\in A &\implies x\in A\cup C\\
&\implies x\in B\cup C\\
x\not\in C,x\in B\cup C &\implies x\in B
\end{align}
$$
In either case, $x\in A \implies x\in B$.  The arguments work the same with $A$ and $B$ switched, so $x\in B \implies x\in A$.  Thus, we conclude $A=B$.
