[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, 10.1.1.7.3029), 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.
