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By Rudolf Carnap and Yeshoua Bar-Hillel

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To illustrate the importance and use of the functions defined in this section, let us work out a different numerical example, albeit an artificially simplified one, for ease of computation. Let h 1 be 'Jones is bright', h and h 3 be 'Jones is dull'. be 'Jones is average (in intelligence)', Somebody who is interested in Jones's intelligence makes him undergo a certain test. Let now k 1 be 'Jones achieves more than 80 percent (in this test)', k 2 be 'Jones achieves between 60 percent and 80 percent', and k 3 be 'Jones achieves less than 60 percent'.

In our case, according to D6, sp(inf, H, K, e) = x 2 + 2Ix -44- 0+ Ix 4 2 = 4' a result that could, of course, also have been obtained from T13. that the test was a pretty poor one. 25 of a unit of information on Jones's intelligence. The difference between 1. 5 and 0. , 1. 25, is the value of est(inf, H/K, e), according to Tll1. D5 or T10. The same value would be obtained by using either We may say that by applying the test instead of measuring the intelligence directly, we must content ourselves with expecting a "loss" of 1.

5. For est(inf, H. 75. This verifies T9. It is obvious that not all h's and k's are inductively independent. To find the various est(inf, H, e. kq), we compute first all c(hp, e. kq). We get c(h1 e. k c(h, e. k 2 ) = c(h 3 , e. k 2 ) (All other c(hp, e. k 2) __ 3) = c(h 3 , ) = 1 4 = 0. ) Hence we have est(inf, H, e. kl) = est(inf, H, e. k 3) = 1, est(inf, H, e. 5. Hence we get, according to D4, sp( inf, e) H, kl, k 3 , e) = sp(inf, H, k 2 , e) = 0. The last result is of special importance. And indeed, if Jones achieves between 60 percent and 80 percent in his test, we are "so klug als wie zuvor", we know exactly as much as we knew before.

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