Prove that if $A \cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$ then $A \subseteq B$ I'm having trouble figuring out what to do in this problem.
I know that if $A \subseteq B$ then $A \cup B = B$ and $A \cap B = A$, but I can't wrap my head around on what to do with the given information to get my result.
 A: If $a \in A$ but $a \notin B$:
$a \in A \cup C \subset B \cup C \implies a \in C$.
Hence, $a \in A \cap C \subset B \cap C \subset B$, contradiction. So $a \in B$.
A: When it comes to doing these sorts of simple set theory problems, I find it is most useful to think explicitly about the few ways these mathematical objects can be constructed and destructed.
When it comes to subsets, we can only really construct and destruct by talking about element membership.
To construct proof of $A\subseteq B$, we need to show that $x\in A \implies x \in B$. That is going to be an inevitability (so long as we opt for a direct/constructive proof). Since this is the case, let us suppose that we have some particular element $x\in A$. Our goal now is just to prove that $x\in B$ using the other information at hand.
Now, the only way we can destruct the other two subset statements we have is to break them down as follows:
$A\cap C \subseteq B \cap C$ destructs to $x\in A\cap C \implies x \in B \cap C$ -- I'll call this statement (1).
Similarly, $A \cup C \subseteq B \cup C$ destructs to $x\in A\cup C \implies x\in B \cup C$, which I will call statement (2).
Since we assumed that $x\in A$, we can construct $x\in A\lor x\in C$ and hence $x\in A\cup C$, and hence invoke statement (2) to conclude that $x\in B\cup C$. 
Then since we have $x\in B\cup C$, we can destruct this to $x\in B\lor x\in C$, and then we must consider two cases.
In the first case, $x\in B$.
Then we have shown that $x\in A \implies x\in B$ and hence $A\subseteq B$ as required, so we're done with this case!
In the second case, $x\in C$. But we already know that $x\in A$, so then we can construct $x\in A\land x\in C$ and hence $x\in A\cap C$. But then by statement (1) from earlier, this gives us that $x\in B\cap C$. This destructs to $x\in B \land x \in C$. So we have shown that $x\in A \implies x\in B$ as required, and we're done with this case too!
It's all about the constructors and destructors.
A: Prove that for any sets $X, C$ you have
$$
X = ((X \cup C) \setminus C) \cup (X \cap C).
$$
A: Case 1. Suppose that $x\in A$ and $x\notin C$. Suppose the inclusion $A\cup C\subseteq B\cup C$ is valid. Note that each of the statements below implies the next.


*

*$(x\in A)$ and $(x\notin C)$ and $(A\cup C\subseteq B\cup C)$

*$(x\in A)$ and $(x\notin C)$ and $((x\in A\cup C)\implies (x\in B\cup C))$

*$(x\in A)$ and $(x\notin C)$ and $((x\in A )\mbox{ or } (x\in C))\implies ((x\in B)\mbox{ or }  (x\in C))$

*$(x\in A)$ and $(x\notin C)$ and $(x\in A)\implies (x\in B)$

*$x\in A$, $x\notin C$ and $A\subseteq B$
Case 2. Suppose that $x\in A$ and $x\in C$. Suppose the inclusion $A\cap C\subseteq B\cap C$ is valid. Note that each of the statements below implies the next.


*

*$(x\in A)$ and $(x\in C)$ and $(A\cap C\subseteq B\cap C)$

*$(x\in A)$ and $(x\in C)$ and $((x\in A\cap C)\implies (x\in B\cap C))$

*$(x\in A)$ and $(x\in C)$ and $((x\in A )\mbox{ and } (x\in C))\implies ((x\in B)\mbox{ and }  (x\in C))$

*$(x\in A)$ and $(x\in C)$ and $(x\in A)\implies (x\in B)$

*$x\in A$, $x\in C$ and $A\subseteq B$
