Generalized factorization, Edition: version 17 Aug 2007 by Grant Larsen

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M* (e)= (6) (s+i) m*(e. h) (s+1 ) ' (s) m*(e) - m*(e) s+2 . (7) (7) s +1 c (h, e) = s cont (h/e) = m( ((1), (6), (7)). (8) x (1 1 ((3), (6), (7)) = m*(e) X s-s+ e ) have To a assume numerical example, We get+2 s=10. To have a numerical example, assume s10. h) = m*(e) X *h, c (h, e) 1+ (11) s +1 12 cont*(h/e) = m*(e) x (12) 11-s 12 (13) inf*(e) = Log i*1 (14) 1 The values given in Table VI are calculated according to these formulas. Table VI s1 m* (e) m (e. h) c*(h, e) cont*(e) cont*(h/e) inf*(e) inf*(h/e) 0 0.

Covering, intuitively speaking, the same ground. It seems to us that the choice of the class of the Q-sentences as the class relative to which the efficiency of a language is defined is a natural one, though it certainly is not the only plausible one. The efficiency of a language, as defined here, changes, as a function of e, with a change of the evidence taken into account. A language *This loose statement is in need of much elaboration. This is expected to be achieved at a later stage. We have in mind that the languages L.

K 2 , ... k m , Then we define -39- --------_ll-----L--(l_l __-I Dl 1-3. m n est(in, H. K, e) =D c(hp. kq, e) x in(hp . kq/e). q=l p=l With respect to the explicatum inf, the following theorem can be proved: T11-9. est(inf, H. K, e),< est(inf, H, e) + est(inf, K, e), where equality holds only if, for all p and q, c(hp. kq, e) = c(hp, e) x c(kq, e), in other words, when the h's and the k's are inductively independent on e (with respect to that m-function on which c is based). H = El11-3. Let e = 'a.

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