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On 5/6/24 2:28 PM, olcott wrote:Sure *D is simulated by H* could mean that *D is never simulated by H*On 5/6/2024 11:19 AM, Mikko wrote:And, since THIS STATEMENT puts no specifications on the design of H, I have shown that your claim is incorrect.On 2024-05-05 17:02:25 +0000, olcott said:>
>The x86utm operating system: https://github.com/plolcott/x86utm enables>
one C function to execute another C function in debug step mode.
Simulating Termination analyzer H simulates the x86 machine code of its
input (using libx86emu) in debug step mode until it correctly matches a
correct non-halting behavior pattern proving that its input will never
stop running unless aborted.
>
Can D correctly simulated by H terminate normally?
00 int H(ptr x, ptr x) // ptr is pointer to int function
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 }
>
*Execution Trace*
Line 11: main() invokes H(D,D);
>
*keeps repeating* (unless aborted)
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.
>
The above execution trace proves that (for every H/D pair of the
infinite set of H/D pairs) each D(D) simulated by the H that this D(D)
calls cannot possibly reach past its own line 03.
When you say "every H/D pair" you should specify which set of pairs
you are talking about. As you don't, your words don't mean anything.
>
Every H/D pair in the universe where D(D) is simulated by the
same H(D,D) that D(D) calls. This involves 1 to ∞ steps of D
and also includes zero to ∞ recursive simulations where H
H simulates itself simulating D(D).
>
If you are going to restrict it to some infinite set built on a specific template, you need to say so, or you are just a liar.Is your memory really that bad?
Of course, that makes you claim much less interesting.--
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