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Problems

A problem is dealt with just as before. If the problem in x is $\neg ( A \rhd B ) $ you find a world y applying lemma 5.2 such that $ x \prec_{\mbox{\footnotesize {B}}} y $, $ A \in y $ and y is such that it is also maximal w.r.t. the $\Box$-inclusion. You then define x ReB y and close off in a minimal way under the $\mathit{ILM_0}$-frame conditions.



Joost Joosten
2000-02-07