Re: Computable Functions --- finite string transformation rules

On 4/25/2025 12:53 AM, olcott wrote:

On 4/24/2025 10:00 PM, dbush wrote:

On 4/24/2025 10:50 PM, olcott wrote:

On 4/24/2025 6:07 PM, Richard Damon wrote:

On 4/24/25 3:41 PM, olcott wrote:

On 4/24/2025 2:12 PM, Fred. Zwarts wrote:

Op 24.apr.2025 om 19:13 schreef olcott:

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HHH correctly determines through mathematical induction that
DD emulated by HHH (according to the finite string transformations
specified by the x86 language) cannot possibly reach its final
halt state in an infinite number of steps.

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No, HHH has a bug which makes that it fails to see that there is only a finite recursion,

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*You are technically incompetent on this point*
When the finite string transformation rules of the
x86 language are applied to the input to HHH(DD)
THIS DD CANNOT POSSIBLY REACH ITS FINAL HALT STATE
not even after an infinite number of emulated steps.

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When the defined finite string trasnsformation rules, thos of the x86 language, are applied to this input, completed with the definitions from Halt7.c as stipulated, we see that DD calls HHH(DD), that it will spend some time emulating DDm then it will

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Correctly determine that DD emulated by HHH can never possibly
reach its final halt state even after an infinite number of
steps are emulated.

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Category error.  The fixed code of algorithm HHH, which is part of the input as you agreed, emulates for a fixed number of steps.  Therefore there is no infinite number of steps emulated by algorithm HHH.
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You are flat out stupid about hypothetical possibilities.
Of every possible HHH/DD pair where DD calls HHH(DD) and
DD is emulated by HHH according to the finite string transformation
rules of the x86 language no DD ever reaches its own final halt state.
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 In other words, you're hypothesizing changing the input.
 Changing the input, hypothetically or otherwise, is not allowed.
 

However, UTM(DD) emulates the same input which includes HHH as part if it for a finite number of steps and reaches a final state.  The instructions emulated by UTM are exactly the same as those emulated by HHH up to the point that it aborts, showing that the fixed code of HHH is incorrect in showing non-halting.

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