I would like to know if my attempted solution to problem 4 of chapter 3.5 from Velleman's book How to Prove It is valid. My solution is different from the one suggested by the book and I want to know if my reasoning is fallacious to avoid repeating it in the future, if that is the case.
The problem is as follows:
Suppose $A\cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$. Prove that $A\subseteq B$.
This is my proof attempt:
Suppose $A\cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$. Let $x$ be arbitrary and suppose $x \in A$. Since $x \in A$ it follows that $x \in A \cup C$. But we know that $ A \cup C \subseteq B \cup C$, so $x \in B$ or $x \in C$. We will consider these cases separately.
- Case $x \in B$: then no further reasoning is needed.
- Case $x \in C$: then $x \in A \cap C$, so since $A\cap C \subseteq B \cap C$ it follows that $x \in B$
Thus $x \in B$. Since $x$ was an arbitrary member of $A$ we can conclude that $A \subseteq B$. $\blacksquare$
And this is the book's solution:
Suppose $x \in A$. We now consider two cases:
- Case $x\in C$. Then $x \in A \cap C$, so since $A\cap B \subseteq B \cap C$, $x \in B \cap C$ and therefore $x \in B$.
- Case $x \notin C$. Since $x \in A$, $x \in A \cup C$, so since $A\cap B \subseteq B \cap C$, $x \in B \cup C$. But $x \notin C$, so we must have $x \in B$.
Thus, $x \in B$, and since $x$ was arbitrary, $A \subseteq B$. $\blacksquare$
So, the book's proof uses a division into cases that are not suggested by any of the assumptions and that I did not think of. Is there any problem with my proof?