, By (1) and (2)
there are finitely many, say k ? 0, representations of ? in the form of a conclusion of the rule of proof (FOR): ? 1 ( ? 1 ? 1 µ 1 ), . . . , ? k ( ? k ? k µ k ) ,
, ¬? ? PDL 0 ; for all m ? N, if 1 ? m ? l, there exists a propositional variable p such that ? l ? ¬? m ( ? m
, ?) ? ? m (? ? (µ m ? ¬p))) ? PDL 0 . First
, ? ? l?1 )?; ¬? ?]?)?¬? / ? S. Since S is closed under the rule of proof (FOR), there exists a propositional variable p such that ([(? ? ? l?1 )?2) and (3)
,
,
We define by induction a sequence (? 0 , . . . , ? k ) of formulas such that for all l ? N, if l ? k, the following conditions are satisfied: ?((? ? ? l )?) ? ? S; ¬? . Obviously, the following conditions are satisfied: ?((? ? ? 0 )?) ? ? S ,
, Third, since ?((? ? ? l?1 )?) ? ? S and ? l?1 ? ¬? ? PDL 0
?) ? ? l (? ? (? l ? ¬p))). Obviously, the following conditions are satisfied: ?((? ? ? l )?) ? ? S; ? l ? ¬? ? PDL 0 ; for all m ? N, if 1 ? m ? l, there exists a propositional variable p such that ? l ? ¬? m ( ? m ((? m ? p) ? ?) ? ? m (? ? ,
f (?) ¬? ? S. Without loss of generality, suppose f (?) contains exactly one test, say ??. Thus, f (?)(??); ¬?? ? ? S. Since there are countably many formulas, there exists an enumeration ? 1 , ? 2 , . . . of the set of all formulas ,
Obviously, f (?)(? 0 ?); ? 0 ? ,
,
or there exists a formula µ such that the following conditions are satisfied: f (?)((? n?1 ? µ)?)the rule of proof (FOR), there exists a propositional variable p such that µ ? ¬?( ? ((? ? p) ? ?) ? ? (? ? ,
, In the former case, let ? n = ? n?1 ? ? n . In the latter case
either f (?)(? n ?)n )? ? ? S, or there exists a formula µ such that the following conditions are satisfied: f (?)(? n ?); the rule of proof (FOR), there exists a propositional variable p such that µ ? ¬?, By Lemma, vol.9 ,
, In the latter case, let ? n = ? n?1 ? µ. Obviously, f (?)(? n ?); ? n ? ? ? S. Finally, the reader may easily verify that T = {PDL 0 + ? n : n ? N} and U = {PDL 0 + ? n : n ? N} are maximal consistent theories such that f (?) ? ker( f (?, the former case, let ? n = ? n?1 ? ? n
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