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[Kriminal99]

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The proof is easily defeated by pointing out that we are assuming that a halting function couldn't just piecewise assign "halts" to itself instead of actually running on itself, which would prevent the existence off any partial function which would attempt to have the halting algorithm run on it's own encoding


Also, Consider the following proof.

Consider a set of Universal Turing Machines that at some point simulate their input machine. Call the set S.

Now we construct another set of machines "G" that takes the encoding of an element of S and copies it, then simulates S on it's self.

G is a subset of S. Therefore G can run on itself. G(G) infinitely loops as defined because it copies it's own input and simulates, then the simulation copies, then simulates, and that simulation copies then simulates etc. forever.

Thus any function G of this sort can not be claimed to be able to do anything BUT loop when run on itself. Thus any proof which depends on the creation and running of such a function on itself and claims that the function retains some ability other than looping is assuming a falsehood. All Halting proofs assume this falsehood, which is the source of the derived contradiction.