Suppose $A \cap C \subseteq B \cap C$ and $A \cup C\subseteq B \cup C$. Prove that $A \subseteq B$ This is Velleman's exercise 3.5.4:
Suppose $A \cap C \subseteq B \cap C$ and $A \cup C\subseteq  B \cup C$. Prove that $A \subseteq B$
This is the proof given by the book (which I understand completely):
Proof. Suppose x ∈ A. We now consider two cases:
Case 1. $x \in C$. Then $x ∈ A \cap C$, so since $A \cap C\subseteq B \cap C, x \in B \cap C$,
and therefore $x \in B$.
Case 2. $x \notin C$. Since $x \in A, x \in A \cup C$, so since $A \cup C\subseteq B \cup C$,
$x \in B \cup C$. But $x \notin C$, so we must have $x \in B$.
Thus, $x \in B$, and since x was arbitrary, $A \subseteq B$.
I was wondering if one could write a proof like this one in below:
Proof. Let x be an arbitrary element of A. Then by $A \cup C \subseteq B \cup C$, we have either $x \in B$ or $x \in C$. Now we consider these two cases:
Case 1. x is an element of B. 
Case 2. x is an element of C. 
Since in one of the cases $x \in B$ and since x was arbitrary, $A \subseteq B$.
 A: One can give a more algebraic proof as follow:
$$
\begin{array}{lcl}
A 
& = & A\cup (A \cap C)\subseteq A\cup(B\cap C) \\
& = & (A \cup B)\cap (A\cup C)\subseteq (A\cup B)\cap (B\cup C) \\
& =& B\cup(A \cap C)\subseteq B\cup (B\cap C) \\
& =& B
\end{array}
$$
A: First, from $A \cap C \subseteq B \cap C$, we have:
$A \cap C \subseteq B \cap C$
$\Rightarrow$ $A \cap C \subseteq B \cap C \subseteq B$
$\Rightarrow$ $A \cap C \subseteq B$
Then, from $A \cup C \subseteq B \cup C$, we have:
$A \cup C \subseteq B \cup C$
$\Rightarrow$ $A \subseteq A \cup C \subseteq B \cup C$
$\Rightarrow$ $A \subseteq B \cup C$
Also, $A \cap C \subseteq B$
$\Rightarrow$ $(A \cap C) \cup (A \setminus C) = A \subseteq B \cup (A \setminus C)$
And, $A \subseteq B \cup C$
$\Rightarrow$ $(A \setminus C) \subseteq (B \cup C) \setminus C = (B \setminus C) \subseteq B$
$\Rightarrow$ $(A \setminus C) \subseteq B$
$\Rightarrow$ $B \cup (A \setminus C) = B$
Therefore, $A \subseteq B \cup (A \setminus C) = B$, which means $A \subseteq B$ is true.
