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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.

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  • $\begingroup$ The sentence A⊆B∧A⊆C→A=∅∨(B∩C≠∅) is not clear. Do you mean instead?: $(A \subseteq B) \land (A \subseteq C) \rightarrow (A = \emptyset) \lor (B \cap C \neq \emptyset)$ $\endgroup$ – Raj Sep 19 '15 at 19:25
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The logic is okay, but the sloppy notation will bite you if you don't start being more careful. I have yet to see someone take notational "shortcuts" like you have (and I've seen lots of people take them), and not eventually confuse themselves (and others) because of it. Careful notation is a halmark of good mathematics.

Particularly, the line

we know that x∈B∧x∈C≡B∩C

As it stands, that sentence is nonsense. What you really mean is

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

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