[This question has been asked on MathOverflow with no luck a month ago.]

Let me first clarify my definitions. For a word $w \in \Sigma^*$, with $\Sigma =\{a_1, \ldots, a_n\}$, I define two structures:

$\mathbb{N}(w) = \langle \mathbb{N}, <, Q_{a_1}, \ldots, Q_{a_n} \rangle$,

and the more usual word model:

$\mathbb{N}^r(w) = \langle \{0, \ldots, |w|-1\}, <, Q_{a_1}, \ldots, Q_{a_n} \rangle$,

where $Q_{a_i} = \{p \mid w_p = a_i\}$.

Then WS1S is the set of second order formulas with models of the form $\mathbb{N}(w)$, with order, and for which second order quantification is limited to finite subsets of the domain. MSO is the set of second order formulas with models of the form $\mathbb{N}^r(w)$, with order.

The usual proof that REG = WS1S proves at the same time that MSO = WS1S. My question is then, for which first or second order relations can we keep this to be true?

For instance, if we add a unary predicate $E(X)$ which says that a (monadic) second order variable contains an even number of objects, we add no power, as $E(X)$ is expressible as "there exists $X_1$ and $X_2$ that partition $X$, in such a way that if an element is in $X_i$ the next one in $X$ is in $X_j$, $i \neq j$, and the first element of $X$ is in $X_1$ and the last is in $X_2$."

Now, if we add a predicate $|X| < |Y|$, then WS1S becomes undecidable (see Klaedtke & Ruess,, while MSO stays trivially decidable.

Thank you.

Edit: As a side question, ... is this question of interest? I mean, I'm no expert in the field, so I'm not sure this question is relevant.

  • 1
    $\begingroup$ We do need Latex math support... $\endgroup$ Commented Aug 16, 2010 at 20:57
  • $\begingroup$ Ain't we just a click away from it? $\endgroup$ Commented Aug 16, 2010 at 21:18
  • $\begingroup$ also see the greasemonkey hack mentioned here: meta.cstheory.stackexchange.com/questions/3/latex-math-support $\endgroup$ Commented Aug 17, 2010 at 7:36
  • $\begingroup$ Was there no joy on MO because they couldn't figure it out, or because they found your question uninteresting? (I'm guessing the latter) $\endgroup$ Commented Aug 18, 2010 at 17:55
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    $\begingroup$ Your model $\mathbb{N}(w)$ is very unusual : what is the need of considering a finite word on an infinite underlying structure ? If you forget the letter predicates (which do not change the problem a lot), you are in fact comparing MSO on finite linear orders, and WS1S on the order $\omega$. This is strange, because it is more intuitive to compare different logic formalisms on the same structure. $\endgroup$
    – Denis
    Commented May 12, 2012 at 18:10


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