Let $S_1,\dots,S_n$ be variables representing unknown sets. A set expression has the form $S_i$, $\overline{E}$ (the complement of $E$), or $E \cap E'$, where $E,E'$ are set expressions. A constraint has the form $E = \emptyset$ or $E \ne \emptyset$, where $E$ is a set expression. Given a conjunction of constraints, the problem is to determine whether they are satisfiable, i.e., whether there is an assignment from variables to sets that makes all of the constraints hold.

Conjecture: if there exists a satisfying assignment, then there is a satisfying assignment where all sets are subsets of a universe $U$ with $k$ elements, where $k$ is the number of constraints of the form $E \ne \emptyset$.

Is this conjecture true?

I suspect it is, but I can't seem to prove it. I can prove it if there is no complement operator, but my proof strategy seems to go awry if there are complements. (Proof strategy: suppose the constraints are $E_1 \ne \emptyset$, .., $E_k \ne \emptyset$, and suppose we have any satisfying assignment. Let $x$ be the smallest element of $E_1 \cup \dots \cup E_k$, where $x$ is an element of $E_i$; then add $x$ to $U$, discard $E_i$, and repeat. Once this is finished, replacing each set $S_i$ with $S_i \cap U$ still leaves all constraints satisfied.) If this conjecture is true, it is a sort of "small-world theorem", which says that a not-too-large universe suffices for checking satisfiability. These kinds of theorems are helpful for improving the performance of constraint-solving algorithms.


1 Answer 1


The conjecture is true, and the proof strategy listed there can be made to work.

Let $s = (s_1,\dots,s_n)$ be a satisfying assignment for the variables that makes all of the constraints hold. We'll write $E(s)$ for the result of evaluating the expression $E$ with the variables $S_i$ set to $s_i$. In the following, define $U$ by the procedure listed in the question: let $E_1 \ne \emptyset$, ..., $E_k \ne \emptyset$ be the constraints of second type; pick any element of $E_1(s) \cup \cdots \cup E_k(s)$, call it $x$; add $x$ to $U$; delete all of the constraints $E_i \ne \emptyset$ such that $x \in E_i$; and repeat. Define $s'_i = s_i \cap U$.

Now I claim that $s'_1,\dots,s'_n$ is also a satisfying assignment that makes all of the constraints hold.

First, let's consider the constraints of the form $E = \emptyset$. I claim that if $E(s) = \emptyset$, then $E(s') = \emptyset$. In particular, I claim that $E(s) \cap U = E(s') \cap U$ for all set expressions $E$. The proof is by recursion: by definition of $s'$, we have $S_i(s) \cap U = S_i(s') \cap U$; also if $t \cap U = t' \cap U$, then $\overline{t} \cap U = \overline{t'} \cap U$; and if $t_1 \cap U = t'_1 \cap U$ and $t_2 \cap U = t'_2 \cap U$, then $(t_1 \cap t_2) \cap U = (t'_1 \cap t'_2) \cap U$. So this means that $s'$ satisfies all of the constraints of the first type (all of the constraints of the form $E = \emptyset$), since $s$ does.

Next, consider the constraints of the form $E \ne \emptyset$. We know $E(s) \ne \emptyset$. Therefore, there was a step in the iterative definition of $U$ when we chose some element $x \in E(s)$ and added it to $U$. We proved earlier that $E(s) \cap U = E(s') \cap U$. Since $x \in E(s)$ and $x \in U$, it follows that $x \in E(s') \cap U$, and thus $E(s') \ne \emptyset$. This means that $s'$ satisfies all of the constraints of the second type, too.

We've shown that if there is any satisfying assignment for all constraints, then there exists a satisfying assignment where all sets are subsets of $U$. This completes the proof of the conjecture.

  • 1
    $\begingroup$ A good question. But you should hide the spoiler just in case someone else wants to solve it (him/her)self. :) $\endgroup$
    – fade2black
    Jul 11, 2017 at 20:46

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