Prove that if $A \subseteq B$ and $A \subseteq C$ then $A \subseteq B \cap C$. Is the proof of this statement correct? Currently in high school—so I'm kinda new to this—and working through "How to Prove It: A Structured Approach"(3rd Edition) by Daniel J. Velleman.


*Prove that if $A \subseteq B$ and $A \subseteq C$ then $A \subseteq B \cap C$.

Scratch work. Quantified: $$(\forall x) [((x \in A \to x \in B) \wedge (x \in A \to x \in C)) \to (( x \in A \to x \in B) \wedge x \in C))]. $$ So, we assume:  $(x \in A \to x \in B)$, $ (x \in A \to x \in C))$ and $ x \in A$, since they're the preceding statements in the conditionals. Hence we have to prove that $x \in B$ and $x \in C$. Since we can suppose that $x \in A$, then, by modus ponens, $x \in B $ and $x \in C$.
Theorem.
Suppose A, B, and C are set. If $A \subseteq B$ and $A \subseteq C$ then $A \subseteq B \cap C$.
Proof.
Let $x$ be arbitrary. Suppose $x \in A$, $A \subseteq B$, and $A \subseteq C$. Because $x \in A$, then by the definition of the subset, $x \in B$, and $x \in C$. Since $x$ was an arbitrary element of $A$; $x \in B$ and $x \in C$, we can conclude that $A \subseteq B \cap C$. Therefore if $A \subseteq B$ and $A \subseteq C$ then $A \subseteq B \cap C$.
 A: I have one quibble with the way you have translated the problem into logical symbols.  You should start by assuming $A \subseteq B$ and $A \subseteq C$, and then you have to prove that $A \subseteq B \cap C$.  Once you have done that, your givens are $A \subseteq B$, which I would write as $\forall x(x \in A \to x \in B)$, and $A \subseteq C$, which I would write as $\forall x(x \in A \to x \in C)$.  Your goal is $A \subseteq B \cap C$, which means $\forall x(x \in A \to x \in B \cap C)$.  So there is a "$\forall x$" at the beginning of each given, and also one at the beginning of the goal.  I wouldn't put a single "$\forall x$" at the beginning of the entire theorem.
Based on the logical form, I would expect the steps of the proof to be:

*

*Assume $A \subseteq B$ and $A \subseteq C$.

*Let $x$ be arbitrary.

*Assume $x \in A$.

*[Reasoning to be filled in here.]

*Therefore $x \in B \cap C$.

*Since $x$ was arbitrary, we can conclude that $A \subseteq B \cap C$.

To complete the proof, fill in step 4.
