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)?

$\endgroup$
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. $\endgroup$ May 9, 2015 at 18:01

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.