4
$\begingroup$

According to this paper, by Vaananen, the $LS$ number for $2^{nd}$ order logic is given by "the supremum of $Π_{2}$-definable ordinals", where "The Lowenheim-Skolem number $LS(L)$ of $L$ is the smallest cardinal $\kappa$ such that if a theory $T\subseteq L$ has a model, it has a model of cardinality $< \max(\kappa, |T|)$."

So I'm wondering if the $LS$ number is larger for higher-order logics (compared to $2^{nd}$ order logic)?

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ Second-order and $n$th order logic have the same LS number: you can emulate $n$th order logic by $n$-sorted 2nd order logic, as you can state in 2nd order logic that every subset of the $i$th sort is represented in the $(i+1)$th sort. $\endgroup$
    – Emil Jeřábek
    Commented May 9, 2015 at 18:01

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .