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May 29, 2010 at 21:49 comment added Adam Okay, I've repaired the question in two ways: first, I've accounted for your clarification on what $\alpha$ actually means (I had it wrong) and secondly I accounted for the fact that ${\mathcal L}^\alpha_{\beta,\gamma}={\mathcal L}^\alpha_{\beta,\lceil\gamma\rceil}$.
May 18, 2010 at 20:39 comment added Joel David Hamkins Sure, I'll look for a revised question from you later. I had a feeling that this wasn't the example for which you were searching.
May 18, 2010 at 20:17 comment added Adam You are absolutely correct, but I think this mostly points to a (serious) defect in how I phrased the question. In light of my error and your very helpful answer to my other question (#25071), let me take a day or two to try to figure out how to rephrase the question to correct this problem. Thank you once again!
May 18, 2010 at 19:22 comment added Joel David Hamkins I see that you want to consider sentences rather than all well-formed formulas of the logic. In this case, consider $\forall x \forall y x=y$ , which is allowed in $L_{\omega,2}^\omega$, but not in $L_{\omega,\infty}^2$.
May 18, 2010 at 19:16 history edited Joel David Hamkins CC BY-SA 2.5
added 1 characters in body
May 18, 2010 at 19:08 history answered Joel David Hamkins CC BY-SA 2.5