having some trouble with proof that if $A \subseteq B$ then $(A \cap C) \subseteq (B \cap C)$ I am really new to proofs, and I don't know why I am having some troubles understanding this and how do they get to the solution to this question:
For all sets $A$ and $B$, if $A \subseteq B$ then $(A \cap C) \subseteq (B \cap C)$
My teacher's explanation:
\begin{align}
\forall x \in A \cap C &\Longleftrightarrow x \in A \wedge x \in C & \text{By definition of intersection} \\
&\Longleftrightarrow x \in B \wedge x \in C & \text{Since } A \subseteq B\\
&\Longleftrightarrow x \in (B \cap C)  & \text{As intersection mandates}
\end{align}
$\therefore A \cap C \subseteq B \cap C$
I asked him more explanation about it, and the only thing he told me it's because $A$ and $B$ are subsets of $C$ thus the whole thing is possible. I kind of understood that, and that $A$ is a subset of $B$. But then standing to what my book says, you have to be clear in math to proof, and I would not say it like the above, because I seriously don't understand it.
I guess my question is, first of all, how comes the above is valid if $A \subseteq B$ then $(A \cap C) \subseteq (B \cap C)$. Second, can I represent it like this if I wanted to:
$$(A \Longrightarrow B) \Longrightarrow ((A \wedge C) \Longrightarrow (B \wedge C))$$
and third how would I prove it?
Thank you so much in advance, sorry for the silly question.
 A: Several things play a role here:


*

*The definition of intersection is extensional, i.e., we define, as always(?) for sets, the intersection by the elements it contains. The intersection $A\cap B$ is thus characterized by the defining property
$$ \forall x\colon (x\in A\cap B\leftrightarrow (x\in A\land x\in B))$$
This definition is used (with different sets) for the first and the third $\iff$ in your proof.

*The definition of "subset" is also extensional, namely $$ A\subseteq B:\leftrightarrow \forall x\colon(x\in A\to x\in B)$$

*for any statements $\phi,\psi,\chi$, the implicatoin $\phi\to\psi$ entails that $(\phi\land \chi)\to(\psi\land \chi)$. Here's a proof (using the rules of natural deduction): We are given that $\phi\to \psi$. Assume $\phi\land \chi$. Then $\phi$. As well as $\chi$. By modus ponens from $\phi\to \psi$ and $\phi$, we get $\psi$. So now $\psi$ and $\chi$. Therefore $\psi\land\chi$. We derived $\psi\land\chi$ on the aassumption $\phi\land\chi$; therefore $(\phi\land\chi)\to(\psi\land\chi)$. In summary, $$\phi\to \psi\vdash (\phi\land\chi)\to(\psi\land\chi).$$


Now we have all formal steps that are required to get from 
$$ x\in A\land x\in C$$
to 
$$ x\in B\land x\in C,$$
namely we have $x\in A\to x\in B$ because we are given that $A\subseteq B$, and then we imply the result from 3 above. Note that this is in fact only an $\implies$, not an $\iff$!

For what it's worth, the given proof has also some formal problems with treating the $\forall x$ (it disappears silently in the middle).
A formally stricter  (while still preferably written mostly in natural language) proof might read like this:
Claim. Let $A,B,C$ be sets. If $A\subseteq B$, then $A\cap C\subseteq B\cap C$.
Proof.
Assume $A\subseteq B$.
By definition, this means
$$\tag1 \forall x\colon (x\in A\to x\in B)$.$$
Let $x$ be arbitrary.
Assume $$\tag2x\in A\cap C.$$
By the definition of $\cap $, this is equivalent to $$\tag3c\in A\land x\in C.$$
Hence we have $x\in A$ and $x\in C$. 
If we specialize $(1)$ to $x$, we obtain $x\in A\to x\in B$. Together with $x\in A$, this implies $x\in B$. From this an d$x\in C$, we conclude $x\in B\land x\in C$. By the definitin of $\cap$, the latter is equivalent to $x\in B\cap C$. As we showed $x\in B\cap C$ from $x\in A\cap C$, we conclude that $x\in A\cap C\to x\in B\cap C$, and as $x$ was arbitrary, we have
$$ \forall x\colon(x\in A\cap C\to x\in B\cap C).$$
By the definition of $\subseteq$, this simply states that
$$ A\cap C\subseteq B\cap C,$$ 
as was to be shown. $\square$
A: The given problem takes the following form:

For all sets $A, B$, and $C,\quad$ $A\subseteq B\implies A\cap C\subseteq B\cap C$.

Proof.(We call this Direct Proof)
Let $A, B,$ and $C$ be sets and assume that $A\subseteq B$. We want to show that  $A\cap C\subseteq B\cap C$. Let $x\in A\cap C$. Then by using the definition of intersection, we get $x\in A$ and $x\in C$. Since $x\in A$ and $A\subseteq B$, it follows from the definition of subset that $x\in B$. So, were able to show that $x\in B$ and $x\in C$. Again, by using the definition of intersection, we conclude that $x\in B\cap C$. Because $x$ was arbitrarily chosen, we conclude that $A\cap C\subseteq B\cap C$. 
A: I'd just like to remark that the arrows in your teachers explanation should not all be double arrows. They should be $\Rightarrow$ instead. The double arrow implies that containment would go both ways but this is only true if the sets are equal.
Your teacher has asked you to prove: For all sets A and B, if $A ⊆ B$ then $(A \cap C) ⊆ (B \cap C)$.
I suppose you could write it using your proposed notation:
Let $P$ be the sentence "x belongs to $A$, $Q$ be the sentence "x belongs to $B$," $R$ be the sentence "x belongs to $C$ and let $A ⊆ B$. Then this can be written:
$P \wedge R \Rightarrow Q \wedge R$
This conditional statement fails if $P \wedge R$ is true and $Q \wedge R$ is false. $P \wedge R$ is true if both $P$ and $R$ are true. Consider, can $Q \wedge R$ be false? Only if either $Q$ or $R$ is false. However we have assumed $R$ to be true and we know $Q$ is true because we have assumed $P$ to be true and $P \Rightarrow Q$.
Therefore, the statement is always true.
A: Ok let me see if I can answer my own question now (I will not give my self credit for it, but if you can disprove it by all means).

If  $A \subseteq B$ then ($A \cap C) \subseteq (B \cap C)$.

Assume $A \subseteq B$. Then by definition of subset, the statement $\forall x (x \in A \Longrightarrow x \in B)$  is true.
We want to show that $(A \cap C) \subseteq (B \cap C)$. This can be done if we can show that the statement $\forall x (x \in A \wedge x \in C) \Longrightarrow (x \in B \wedge x \in C)$ is true.
Let $p,q,r$ stands for the propositions $x \in A, x\in B$ and $x \in C$. Then we get $p\implies q$. Now, we want show that the statement $(p \wedge r) \Longrightarrow (q \wedge r)$ is true.
This statement will be false, if and only if $q \wedge r$ is false which is when either p or r is false and true or both are false. We know that q is true or false because we assumed that $p \Longrightarrow q$ is true, thus only if both p and q are true or false or if only p is false the statement is true. Hence the statement will be true either case, because if $q \wedge r$ is false then r must be, because if q is false $p \Longrightarrow q$ will not be true, and if r is false $p \wedge r$ will be false and $q \wedge r$ also making the statement still  true.
$\therefore$ If $A \subseteq B$ then $(A \cap C) \subseteq (B \cap C)$ is a true statement.
