Deficiencies

Again the deficiencies are harder to deal with. A first difficulty you encounter is in incorporating the $\mathit{ILM_0}$-frame condition. So, if you eliminate a deficiency in x w.r.t. y by defining y S_x y', you should make sure that every possible successor of y' can also be a successor of some intermediate x' (intermediate means x R x' R y). This is done, as stated before, by demanding $x' \subset_{\Box} y'$. Lemma 5.4 tells us that this is indeed possible. So for every y''with $y' \prec y''$, you immediately have $x' \prec y''$. A closer examination brings us to the second difficulty, that is, that there might be more intermediate worlds between x and y. But the invariant 6 tells us that there is always an intermediate world that is maximal w.r.t. the $\Box$-inclusion relation. This settles the second difficulty if we just apply lemma 5.4 every time to the $\Box$-maximal intermediate world. A third difficulty is found when realizing that the S-relation must be transitive. Say $C \rhd D$ is a deficiency in x w.r.t. y and we have $x \prec_B x' R y$. We want to create a B-critical successor y' of x with $D \in y'$ and $x' \subset_{\Box} y'$. By doing so, every successor of y' can automatically be defined a successor of x' and the $\mathit{ILM_0}$-frame condition is satisfied. Invariant 6 tells us that without loss of generality we can take this x' to be maximal w.r.t. the $\Box$-inclusion relation. Creating y' containing D, to eliminate the deficiency $C \rhd D$ in x w.r.t. y is done in such a way that $x' \subset_{\Box} y'$. The existence of this required entity is guaranteed by lemma 5.4. Automatically we now have $x R x'' R y \rightarrow x'' \subset_{\Box} y'$. We accordingly set y S_x y'. The general frame conditions demand x R y'. Now it may happen that $C' \rhd D'$ is a deficiency in xw.r.t. y'. It is easy to find a world y'' with D' in it and being B-critical above x. The transitivity of S_x forces us to also have y S_x y'', but by no means we can ensure that $x' \subset_{\Box} y''$. Somehow we have to relate the possible deficiencies to each other. As we have only finitely many relevant sentences of the form $C \rhd D$, the following lemma helps us out.

Lemma 5.5  

$$\begin{array}{cll} \par E_0 \rhd F_0 , \ldots , E_n \rhd F_n ... ...} \vdots& \vdots \\ & \vee ( F_n \wedge \Box C ). \end{array}$$

As the difference between the various E_i's is not essential, one has the lemma for any permutation of the indices. If one encounters a deficiency in x w.r.t. y, say $E_0 \rhd F_0$, one proceeds as follows. First a list is made of all relevant sentences of the form $C \rhd D$. Let this list be $E_0 \rhd F_0 , \ldots , E_n \rhd F_n$. You can consider these as the possible deficiencies. Let x' be such that x R x' R y and $x R x'' R y \rightarrow x'' \subset_{\Box} x'$. All deficiencies in x w.r.t. y will be eliminated in such a way that no new deficiencies will occur in x w.r.t. the newly created worlds. Again this process shall henceforth be referred to as the process of making an S_x^y-block. The $\Box$-inclusion of x' shall be taken into account while creating this S_x^y-block.

As stated before lemma 5.5 actually represents a manifold of statements. All of them will be used in eliminating the deficiency $E_0 \rhd F_0$. Let $\pi$ be a permutation of $\{ 1,2,\ldots ,n \}$. With any $\pi$ a series of sets $\Delta_i^{\pi}$ is defined. $( i \in \{ 0,1, \ldots ,n \}.)$

$$\Delta_0^{\pi} = \{ E_1, \ldots , E_n \} ,\ \ \Delta_{i+1}^{\pi} = \Delta_i^{\pi} \setminus \{ E_{\pi(i+1)} \} .$$

For the sake of convenience we define $\forall \pi : \pi(0) = 0$. The notation $\neg \Delta^{\pi}_i$ stands for the sentence $\bigwedge \hspace{-2.5mm} \bigwedge_{E_j \in \Delta^{\pi}_i } \neg E_j$. With this terminology one can easily write down the n! useful variants of lemma 5.5.

$$P_{\pi}:= \Diamond E_0 \wedge \Box C \rhd {\bigvee \hspace{-... ...0}^n ( F_{\pi(i)} \wedge \neg \Delta^{\pi}_i \wedge \Box C ).$$

For any $P_{\pi}$ lemma 5.3 can be applied. Recall our situation, x R_e^B x' R y with $E_0 \rhd F_0 , \ldots , E_n \rhd F_n$ the possible deficiencies in x w.r.t. y. So applying lemma 5.3 to any $P_{\pi}$ gives a maximal $\mathit{ILM_0}$-consistent set containing one of the disjuncts of the consequent of $P_{\pi}$. Note that $P_{\pi} \in x$ because $x \vdash P_{\pi}$. We now form a set of disjuncts $\mathcal{D}$ as follows. For every $P_{\pi}$ you choose the leftmost disjunct of the consequent which is realizable by applying lemmas 5.3 and 5.5.

Lemma 5.6   Consider the above situation. Let $F_{i_1}, \ldots , F_{i_k}$ be all the formulas that do not occur in any of the n! disjuncts in $\mathcal{D}$, and let F_j be an arbitrary formula of the possible F's which does occur in some disjunct in $\mathcal{D}$. There must be some disjunct in $\mathcal{D}$ where both F_j and $\neg E_{i_1}, \ldots , \neg E_{i_k}$ hold.