Not Homework!

This is exercise 1-5.7(B) from Stoll, Robert R.. Set Theory and Logic (Dover Books on Mathematics) (Kindle Location 787). Dover Publications. Kindle Edition.

Prove that an equation in $\mathcal{X}$ with righthand member $\emptyset$ can be reduced to one of the form $\left(\mathcal{A}\cap\mathcal{X}\right)\cup\left(\mathcal{B}\cap\mathcal{\overline{X}}\right)=\emptyset$. (Suggestion: Sketch a proof along these lines. First, apply the DeMorgan laws until only complements of individual sets appear. Then expand the resulting lefthand side by the distributive law 3 so as to transform it into the union of several terms $\mathcal{T}_{i}$, each of which is an intersection of several individual sets. Next, if in any $\mathcal{T}_{i}$ neither $\mathcal{X}$ nor $\mathcal{\overline{X}}$ appears, replace $\mathcal{T}_{i}$ by $\mathcal{T}_{i}\cap\left(\mathcal{X}\cup\mathcal{\overline{X}}\right)$ and expand. Finally, group together the terms containing $\mathcal{X}$ and those containing $\mathcal{\overline{X}}$ and apply the distributive law 3.)

I'm not asking for a solution.

But, if I understand the "suggestion", he wants me to do this with some kind of formal expression that represents the most general set theoretical equation in one variable. I'm not sure exactly how to formulate such an expression.

What comes to mind is something that looks kind of like $$\left(\mathcal{X}\cap\left(\cap_{i}\mathcal{A}_{i}\right)\cap\left(\cap_{i}\overline{\mathcal{\mathcal{B}}}_{i}\right)\right)\cup\left(\overline{\mathcal{X}}\cap\left(\cap_{i}\mathcal{C}_{i}\right)\cap\left(\cap_{i}\overline{\mathcal{D}}_{i}\right)\right), etc.$$

Is there some form of set theoretical equation that is general enough to use for this proof?

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An allowed expression can be the variable $X$, a constant $A_i$, or something of the form $A \cup B$, $A \cap B$, or $\overline{A}$, where $A$ and $B$ are allowed expressions.

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  • $\begingroup$ I haven't figured it out yet, but in the deep recesses of forgotten knowledge I recognize that you have given me the key. $\endgroup$ – Steven Thomas Hatton Dec 30 '17 at 18:26

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