# Is my informal proof for $A\subseteq B\wedge A\subseteq C\implies A=\varnothing \vee(B\cap C\ne \varnothing)$ valid?

I am new to proofs, so I just want to make sure that my informal proof makes sense.

Original problem: Prove or Disprove $$A\subseteq B\wedge A\subseteq C\implies A=\varnothing \vee(B\cap C\ne \varnothing)$$

Informal Proof: Let $x$ be an arbitrary element of $A$.

It follows from $A\subset B$ that $x$ must also be an element of $B$. It also follows from $A\subseteq C$ that $x$ must also be an element of $C$.

From both of these conclusions, we know that $x\in B\wedge x\in C = B\cap C$ by definition of intersection. Therefore, if $A$ contains any elements, then the consequent $B\cap C\ne\varnothing$ holds true.

The only other case would be if $A=\varnothing$ which is also supported by our consequent. Thus, we have shown $A\subseteq B\wedge A\subseteq C\implies A=\varnothing \vee (B\cap C\ne \varnothing )$.

Q.E.D

Thanks.

• The sentence A⊆B∧A⊆C→A=∅∨(B∩C≠∅) is not clear. Do you mean instead?: $(A \subseteq B) \land (A \subseteq C) \rightarrow (A = \emptyset) \lor (B \cap C \neq \emptyset)$ – Raj Sep 19 '15 at 19:25

"so $x \in B$ and $x \in C$. Therefore $x \in B \cap C$, which is not empty."