Re: Can D simulated by H terminate normally? --- Message_ID Provided

On 5/18/24 3:24 PM, olcott wrote:

On 5/18/2024 2:15 PM, Richard Damon wrote:

On 5/18/24 2:54 PM, olcott wrote:

On 5/1/2024 7:10 PM, Richard Damon wrote:

On 5/1/24 12:11 PM, olcott wrote:

Until you refine your current non-existant definitions of the terms, you have the problem described.
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I can't have any idea what you are saying until you fill in
all of the details of your baseless claims.
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But you refuse to listen.
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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 decided 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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First of all the code template that I am currently referring
has nothing to do with any decider, it only pertains to a
simulator where H correctly simulates 1 to ∞ steps of D
of each H/D pair specified by the following template.

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And that is exactly what my H is. It will simulate all of the steps of D, the D that call that H, till it reaches the end.
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typedef int (*ptr)();  // ptr is pointer to int function
00 int H(ptr x, ptr y);
01 int D(ptr x)
02 {
03   int Halt_Status = H(x, x);
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 }
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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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And mine emulates ALL of them to the final return on line 06

 You have already agreed that this is impossible for pure simulator
H many times. I am stopping at your first big mistake.
 

So, you admit that you are not looking at what I have done, this means that any denials of the ability for my method are just admitted LIES and proof that you are just a pathological liar that doesn't even care what is true.