XPost: sci.logic

On 5/9/24 11:10 PM, olcott wrote:

On 5/9/2024 9:31 PM, Richard Damon wrote:

On 5/9/24 11:38 AM, olcott wrote:

On 5/8/2024 8:38 PM, immibis wrote:

On 8/05/24 21:05, olcott wrote:

On 5/8/2024 10:13 AM, Mike Terry wrote:

On 08/05/2024 14:01, olcott wrote:

On 5/8/2024 3:59 AM, Mikko wrote:

On 2024-05-07 19:05:54 +0000, olcott said:

On 5/7/2024 1:54 PM, Fred. Zwarts wrote:

Op 07.mei.2024 om 17:40 schreef olcott:

On 5/7/2024 6:18 AM, Richard Damon wrote:

On 5/7/24 3:30 AM, Mikko wrote:

On 2024-05-06 18:28:37 +0000, olcott said:

On 5/6/2024 11:19 AM, Mikko wrote:

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).

"In the universe" is not a set. In typical set theories >>>>>>>>>>>>> like ZFC there
is no universal set.

This template defines an infinite set of finite string H/D >>>>>>>>>>> pairs where each D(D) that is simulated by H(D,D) also calls >>>>>>>>>>> this same H(D,D).

These H/D pairs can be enumerated by the one to ∞ simulated >>>>>>>>>>> steps of D and involve zero to ∞ recursive simulations of H >>>>>>>>>>> simulating itself simulating D(D). Every time Lines 1,2,3 are >>>>>>>>>>> simulated again defines
one more level of recursive simulation.

1st element of H/D pairs 1 step  of D  is simulated by H >>>>>>>>>>> 2nd element of H/D pairs 2 steps of D are simulated by H >>>>>>>>>>> 3rd element of H/D pairs 3 steps of D are simulated by H >>>>>>>>>>>
4th element of H/D pairs 4 steps of D are simulated by H >>>>>>>>>>> this begins the first recursive simulation at line 01

5th element of H/D pairs 5 steps of D are simulated by
next step of the first recursive simulation at line 02

6th element of H/D pairs 6 steps of D are simulated by
last step of the first recursive simulation at line 03

7th element of H/D pairs 7 steps of D are simulated by H >>>>>>>>>>> this begins the second recursive simulation at line 01

Is this the definition of the infinite set of H? We can think >>>>>>>>>> of many more simulations that only these.

This template defines an infinite set of finite string H/D
pairs where
each D(D) that is simulated by H(D,D) also calls this same H(D,D). >>>>>>>>>
No-one can possibly show one element of this set where D(D)
reaches
past its own line 03.

If H is a decider of any kind then the D build from it reaches >>>>>>>> its line
4 as numberd above. Whether the simulation of D by H reaches
that line
is another question.

*My fully operational code proves otherwise*

I seems like you guys don't have a clue about how infinite
recursion works. You can run the code and see that I am correct. >>>>>>>
I have one concrete instance as fully operational code.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
line 555 u32 HH(ptr P, ptr I) its input in on
line 932 int DD(int (*x)())

HH is completely broken - it uses a global variable which is
allows HH to detect whether it is the outer HH or a nested
(simulated) HH. As a result, the nested HH behaves completely
differently to the outer HH - I mean /completely/ differently: it
goes through a totally separate "I am called in nested mode" code
path!

The encoding of HH is not the pure function that it needs to be to
be a computable function.

*Maybe you can settle this*

The disagreement is entirely over an enormously much simpler thing.
The disagreement is that Richard says that a D simulated by H could
reach past its own line 03 and halt.

Here's the proof:

1. A simulation always produces an identical execution trace to the
direct execution.

*When pathological self-reference is involved this is counter-factual*
That no one can possibly show the steps of how D simulated by H possibly >>> reach line 06 of H proves this.

Richard tried to get away with D never simulated by H as an example
of D simulated by H:

Nope, you are looking at the WRONG message, and I have told you this
multiple times.

Message-ID: <v0ummt$2qov3$2@i2pn2.org>
*When you interpret*
On 5/1/2024 7:28 PM, Richard Damon wrote:

On 5/1/24 11:51 AM, olcott wrote:

*Every D simulated by H that cannot possibly*
*stop running unless aborted by H*

as *D NEVER simulated by H*

you have shown a reckless disregard for the truth
that would win a defamation case.

My H simulated 0 steps of D, of which was ALL of the steps it simulated correctly.

It used logic just as good as your H in determining that it was correct.

You don't seem to understand that logic.

The arguement still holds if H simulates exactly one step (to allow the
claim additional claim of some steps correctly simulated as well as the
claim tha ALL steps were correctly simulated).

Note, you haven't defined what "Simulated" means, and thus it doesn't
need to mean exact reproduction of the steps used in the procedure, so
your claim that my program was D NEVER simulated isn't actually proven.

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