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**james121515** Hi guys!

I just worked on another simple set theory proof, and have come out with what seems like it might work alright, but while I was writing this I kind of felt like I was making some statements that perhaps needed to be more better justified. For example, in the second sentence of the proof, it kind of feels like I just pulled that out of thin air without justification.

$\displaystyle \text{\underline{Theorem}:}$ If $\displaystyle A\subseteq B$, then $\displaystyle \bar{A}= A$\$\displaystyle B \cup \bar{B}$

$\displaystyle \text{\underline{Proof}:}$ To show the left to right containment, let $\displaystyle x \in \bar{A}$. Then $\displaystyle x\notin A$. Note that there are two cases for $\displaystyle B$, either $\displaystyle x \in B$ or $\displaystyle x \notin B$. If $\displaystyle x \in B$, then we have $\displaystyle x \in B$ and $\displaystyle x \notin A$, which implies that $\displaystyle x \in B$\$\displaystyle A$. If $\displaystyle x \notin B$, then $\displaystyle x \in \bar{B}$, and so $\displaystyle x \in B$\$\displaystyle A$ $\displaystyle \cup $$\displaystyle \bar{B}$. Therefore, $\displaystyle \bar{A} \subseteq B$\$\displaystyle A \cup \bar{B}$. To show the reverse containment, let $\displaystyle x \in A$\$\displaystyle B \cup \bar{B}$. Then $\displaystyle x \in B$\$\displaystyle A $ or $\displaystyle x \in \bar{B}.$ So we have two cases to consider. Suppose $\displaystyle x\in B$\$\displaystyle A$. Then $\displaystyle x \in B$ and $\displaystyle x \notin A$. Since $\displaystyle x \notin A$, $\displaystyle x \in \bar{A}$. Suppose $\displaystyle x \in \bar{B}$. Then $\displaystyle x \notin B$. Note that by hypothesis $\displaystyle A \subseteq B$, and thus $\displaystyle x \notin B$ clearly implies $\displaystyle x \notin A$. And so $\displaystyle x \in \bar{A}.$. Since in either case $\displaystyle x \in \bar{A}$, we can conlude that $\displaystyle B$\$\displaystyle A \cup \bar{B} \subseteq \bar{A}$. Since we've shown both containments hold true, we can conlude that $\displaystyle \bar{A}= B$\$\displaystyle A \cup \bar{B}$.

Thanks for taking your time to read this,

James