![]() Alternatively, we could have just shown that must be infinite based on the assumption of (a). ![]() In particular, the very first step tests whether or is a tautology, and if not, then there is at least one element in. A Mathematical Introduction to Logic This Page Intentionally Left Blank A Mathematical Introduction to Logic Second Edition Herbert B. A mathematical introduction to logic Academic Press Herbert Enderton, Herbert B. Also, importantly enough, if does not imply either or, then there is at least one element of which was not yet enumerated, and, therefore, the procedure will not stuck at finding. ComputabilityandProvability 71 §4.1. According to the compactness theorem, if, then there is a finite subset of that also tautologically implies, in particular, every such that will tautologically imply, hence, our procedure will eventually return either “yes” or “no” (once we have enumerated all the members of for either or ). viii Contents §3.3.TheSumofFourSquares 53 §3.4.TheBrahmaguptaPellEquation 55 FurtherReading 67 Exercises 67 Chapter4. If the answer of the first test is “yes” then we know that, if the answer of the second test is “yes” then we know that, otherwise we increase by and continue testing. Here we assume that ’s are defined as long as the procedure returns at least elements (it may run forever after all of the elements of are listed). ![]() (a) Let be the first entries produced by the procedure that enumerates, including.
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