Is the form and correctness of my elementwise proof of this correct? I don't have any other way of getting feedback for my proofs and I want to improve.

Proof. Suppose $A, B, C, D$ are sets such that $A \subseteq C$ and $B \subseteq D$ and let $x \in A \cap B$. It has to be shown that $x \in C \cap D$.

$x \in A \cap B$ means that $x \in A$ and $x\in B$. Because $A \subseteq C$, $x \in C$ and because $B \subseteq D$, $x \in D$. Thus, $x \in C \cap D$.

Thus, if $A \subseteq C$ and $B \subseteq D$, then $A \cap B \subseteq C \cap D$.

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    $\begingroup$ This is excellent. $\endgroup$ – Brian M. Scott Oct 21 '12 at 15:18
  • $\begingroup$ Thanks! I just fixed an error that I made in the title. Should I elaborate more on where $x \in A \cap B$ comes from? It comes from $A \cap B \subseteq C \cap D$, correct? $\endgroup$ – Brandon Amos Oct 21 '12 at 15:22
  • $\begingroup$ @BrianM.Scott Realized that too. Deleted my comment before seeing you reply. $\endgroup$ – hwhm Oct 21 '12 at 15:23
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    $\begingroup$ No need to say any more: the reason for choosing $x\in A\cap B$ initially is clear just from the inclusion that you’re trying to prove. $\endgroup$ – Brian M. Scott Oct 21 '12 at 15:24
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    $\begingroup$ You’re very welcome, and I agree with what Asaf wrote in the answer below. $\endgroup$ – Brian M. Scott Oct 21 '12 at 15:37

This is a very well written proof. You state your assumptions and what you wish to prove, then you use the definitions to prove that.

There is nothing more to add, and nothing to reduce. Incidentally today I had the first class of the semester and this is exactly what I tried to teach my students. If they all write such proofs by the end of the month, I should be proud of my work.


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