Re: Every D correctly simulated by H cannot possible reach its own line 06 and halt

On 5/20/24 2:03 PM, olcott wrote:

On 5/20/2024 6:24 AM, Richard Damon wrote:

On 5/19/24 11:22 PM, olcott wrote:

On 5/19/2024 10:11 PM, Richard Damon wrote:

On 5/19/24 10:52 PM, olcott wrote:

On 5/19/2024 8:10 PM, Richard Damon wrote:

On 5/19/24 8:06 PM, olcott wrote:

On 5/1/2024 7:10 PM, Richard Damon wrote:
>
typedef int (*ptr)();  // ptr is pointer to int function
00 int H(ptr p, ptr i);
01 int D(ptr p)
02 {
03   int Halt_Status = H(p, p);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 int main()
10 {
11   H(D,D);
12   return 0;
13 }
>
In the above case a simulator is an x86 emulator that correctly emulates at least one of the x86 instructions of D in the order specified by the x86 instructions of D.
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This may include correctly emulating the x86 instructions of H in the order specified by the x86 instructions of H thus calling H(D,D) in recursive simulation.
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For every H/D pair of the above template D correctly simulated by
*pure function* H cannot possibly reach its own final state at
line 06 and halt.
>

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Ok, so adding that H is a pure function, that means that since your outer H(D,D) is going to return 0, all logic must be compatible with the fact that EVERY call to H(D,D) will also eventually return 0.
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Remember also, THIS D is defined to call THIS H, that does exactly the same as the H that is deciding it.
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OK, good.

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Right, so it doesn't matter what any other D does, it matters what THIS D does, and this D calls aths H.
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Remember, you reinstated the Computation model by enforcing Pure Functions.
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<snip so that Message ID links to whole message>
We can use my unique time/date stamp as an alternative.
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Remember, YOU are the one saying you are needing to change the definition from the classical theory, where we have things well defined.
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YOU have decider that H is just whatever C code you want to write for it, and D is the input proved. (which doesn't actually match the Linz or Sipser proof, but fairly close).
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With THAT set of definitions we have a lot of options that break your incorrectly assumed results.
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The first method has been discussed here by Flibble. While the final answer he got to doesn't fit the requirements, the first part of the method DOES show that it is possible for an H to simulate to past line 3.
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THe basic idea is that if H(M,d) finds that its simulation of M(d) get to a call to H(M,d) then rather that your idea of just saying it will get stuck and declair the input invalid, since there ARE a number of possible inputs that there is a "correct" answer that H can give to

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That D is calling H does not prove recursive simulation.
That D is calling H with its same parameters does seem
to prove non-halting recursive simulation.

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Nope. Try to actuall PROVE it.
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That is off-topic for this post.
All that we need know is that no D simulated by any H
ever reaches its own line 06 and halts.

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Nope. Make a claim, you need to prove it.
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*In other different post not this one*
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I am using categorically exhaustive reasoning that can work
through every possibility that can possibly exist in a feasible
amount of time as long as the category is very very narrow.

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But you can't PRECISELY define the category, or what you want to reason about, so your logic is worthless as it is baseless.
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 *POINT TO ANY ACTUAL MISTAKE OR AMBIGUITY WITH THIS VERSION*
 typedef int (*ptr)();  // ptr is pointer to int function
00 int H(ptr p, ptr i);
01 int D(ptr p)
02 {
03   int Halt_Status = H(p, p);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 int main()
10 {
11   H(D,D);
12   return 0;
13 }
 In the above case a simulator is an x86 emulator that correctly emulates at least one of the x86 instructions of D in the order specified by the x86 instructions of D.
 This may include correctly emulating the x86 instructions of H in the order specified by the x86 instructions of H thus calling H(D,D) in recursive simulation.
 Execution Trace
Line 11: main() invokes H(D,D);
 keeps repeating (unless aborted)
Line 01:
Line 02:
Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)
 Simulation invariant:
D correctly simulated by H cannot possibly reach past its own line 03.
 For every H/D pair of the above template D correctly simulated by pure function (thus computable function) H cannot possibly reach its own final state at line 06 and halt.
 

Which thus doesn't correct simulate the call to H (since H proves that a call to itself as H(D,D) WILL return the value 0)
Since you have an incorrect step in your simulation, you arguement is just unsound.
Note, a correct simulation of a call to H(D,D) is NOT another simulation of D(D), as H SIMULATES its input, not run it, and even if the "simulation" is "debug stepping", the H still retains control, so if H will abort its simulation at some point, such simulation is not "infinite", even if the out H gives up before it gets to that point.