Does faster exact algorithm for counting independent sets in comparability graphs than general graph exisits? Sorry for not-precise question. :-(

There are several papers concerning exact counting (maximum) independent sets in general graphs. Actually, they concerns counting of solutions of 2SAT. The best of them is $O(1.23^n)$. But the algorithms do not use the specific information of comparability graphs. 

So I wonder whether there exists more powerful(faster) exact algorithm for counting independent sets in comparability graphs?
Bipartite graphs are comparability graphs, and it's known (since Provan and Ball) that counting the number of independent sets in bipartite graphs is #P-complete.

J. S. Provan and M. O. Ball. The complexity of Counting Cuts and of Computing the Probability that a Graph is Connected.  SIAM J. Comput. 12 (1983) 777-788.

If I haven't mistaken, this should answer your question.