Halting theorem

Are there any noncomputable numbers They are certain to exist since the cardinality of the set of all possible programs (which is equal to that of the integers) is less than the cardinality of the set of reals. It is, in fact, easy to construct a noncomputable number, using Turing s Halting Theorem number, in some arbitrary manner, all possible programs that run with, say, a single input instruction Pi (-> l,p2 2,...,pn n,..., and set ( = n2, where q = 1 if the... 

Deterministic Unpredictability Halting Theorem Godel s Theorem... 

THEOREM 3.3 (1) A scheme P is always halting if and only if every consistent... 

THEOREM 3.12 If P is an always halting scheme we can construct a strongly equivalent free tree program scheme. 

THEOREM 6.12 Let Q be any scheme Which halts under all interpretations. The following properties of schemes P and R are partially decidable although not decidable ... 

We have already seen (in section B of Chapter III) that (7) is partially decidable and that whenever (7) holds for P we can convert P into an always halting tree scheme and decide strong equivalence and total equivalence (Theorems 3.6 and 3.8 and Corollaries 3.7 and 3.9). Hence (7) - (11) are partially decidable. 

THEOREM A-l Given an always halting program scheme P, we can construct a quantifier-free wff p such that P is totally correct with respect to TRUE and P ... 

Theorem 7.9 shows that In a gTS all loop, p-a and end failures can be eliminated and the proof provides a procedure for constructing the reduced gTS. From a practical point of view, a recognition schema which Is "reduced" seems to be desired since the program will always terminate and it will never loop or halt before coming up with the final answer. 

Notice that the emptiness problem for LBC (which is equivalent to the halting problem for multicounter machines) is known undecidable. According to Theorem 4, for (single) RBCS, the emptiness problem (i.e., Lang(M, L) = 0 for a given RBCS M and regular bond type L) is undecidable as well. Hence,... 

Second, specific reasons must exist in order to believe that the proposed process is incomputable. Since solving the halting problem is known to be incomputable and adding axioms is incomputable by definition (otherwise they would be theorems), then specific evidence indicates that the proposed process is incomputable. 

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