this is an exercise from How to solve it by Daniel J. Velleman. My proof differs from the one given in the book and I was wondering if it is still valid.

Suppose $A \cap C \subseteq B \cap C$ and $A \cup C \subseteq B \cup C$. Prove that $A \subseteq B$.

My proof: Let $x$ be arbitrary. Suppose $x \in A$. Since $A \cup C \subseteq B \cup C$, then $x \in B \cup C$. Then we have the following two cases.

Case 1: Suppose $x \in B$. This is our goal.

Case 2: Suppose $x \in C$. Since $A \cap C \subseteq B \cap C$, then $x \in B \cap C$, which means $x \in B \land x \in C $, which means $x \in B$.

Since x was arbitrary then $\forall x: x \in A \rightarrow x \in B$, which means $A \subseteq B$.

I don't see any logical errors in my proof, but I'd be really grateful if somebody could check it.

  • $\begingroup$ Looks good to me. $\endgroup$ – R_D Sep 28 '15 at 15:41
  • $\begingroup$ You first suppose that $A \cap B \subseteq B \cap C$ but later use that $A \cap C \subseteq B \cap C$; is the former a typо? $\endgroup$ – Zoran Loncarevic Sep 28 '15 at 15:44
  • $\begingroup$ It's a typo! Thanks a lot of pointing it out. $A \cap C \subseteq B \cap C$ $\endgroup$ – geomquestion Sep 28 '15 at 15:47
  • $\begingroup$ Yes, the proof is now OK. $\endgroup$ – Zoran Loncarevic Sep 28 '15 at 15:54
  • $\begingroup$ Great. Thanks a lot! $\endgroup$ – geomquestion Sep 28 '15 at 16:03

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