, Assume C is consistent, hence a subset of a rational constitution C 2 T. We show that CjS C
, It su¢ ces to specify a constitution C such that CjS violates R. Simply let C be any constitution such that F C. Since F C and C CjS, we have F CjS. So the revised constitution CjS violates R. = ?, it su¢ ces to prove Theorem 4. Consider any theory T. (a) By de?nition, each conditional-completeness requirement of T is conditional on some set P of states, and has at least one U-state. For each such requirement, ?x an arbitrary member m U of U , and de?ne the rule r U = (P; m U ). Let S be any reasoning system containing one such rule for each conditional-completeness requirement. Clearly, S achieves all conditional-completeness requirements of T. (b) Consider any conditional-completeness requirement of T , de?ned by some P and U (where U 6 = ?). Suppose some consistency-preserving reasoning system S achieves this requirement. Since the requirement is achieved, there must be some m 0 2 U such that m 0 2 P jS. To complete the proof, it su¢ ces to show that m 0 is not falsi?able given P. To that end, we consider a consistent C 0 P , and must show consistency of C 0, The ?rst is that C is closed under S, because it is rational and hence in particular satis?es all closedness requirements of T
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