We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the behavior of DDD, which continues per the definition of the x86 instruction set until
the COMPLETE emulation (or direct execution) reaches a terminal
instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and aborted
its emulation on a non-terminal instrucition only have partial knowledge
of the behavior of their DDD, and by returning to their caller, they establish that behavior for ALL copies of that HHH, even the one that
DDD calls, which shows that DDD will be halting, even though HHH stopped
its observation of the input before it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a finite
number of steps WILL halt becuase it will reach its final return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY case,
when we look at the behavior past that point, which DOES occur per the definition of the x86 instruction set, as we have not reached a
"termial" instruction that stops behavior, will see the HHH(DDD) that
DDD called continuing to simulate its input to the point that this one
was defined to stop, and then returns 0 to DDDD and then DDD returning
and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH does
of the behavior of DDD by its PARTIAL emulation with the ACTUAL FULL behavior of DDD as defined by the full definition of the x86 insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are shown
to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because THEIR
DDD will halt. and the HHH that doen't abort never get around to
rejecting its DDD as non-halting.
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of
the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the behavior
of DDD, which continues per the definition of the x86 instruction set
until the COMPLETE emulation (or direct execution) reaches a terminal
instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have partial
knowledge of the behavior of their DDD, and by returning to their
caller, they establish that behavior for ALL copies of that HHH, even
the one that DDD calls, which shows that DDD will be halting, even
though HHH stopped its observation of the input before it gets to that
point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a finite
number of steps WILL halt becuase it will reach its final return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY
case, when we look at the behavior past that point, which DOES occur
per the definition of the x86 instruction set, as we have not reached
a "termial" instruction that stops behavior, will see the HHH(DDD)
that DDD called continuing to simulate its input to the point that
this one was defined to stop, and then returns 0 to DDDD and then DDD
returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH does
of the behavior of DDD by its PARTIAL emulation with the ACTUAL FULL
behavior of DDD as defined by the full definition of the x86
insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around to
rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
On 7/12/24 7:19 PM, olcott wrote:
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of
the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the behavior
of DDD, which continues per the definition of the x86 instruction set
until the COMPLETE emulation (or direct execution) reaches a terminal
instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have
partial knowledge of the behavior of their DDD, and by returning to
their caller, they establish that behavior for ALL copies of that
HHH, even the one that DDD calls, which shows that DDD will be
halting, even though HHH stopped its observation of the input before
it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a finite
number of steps WILL halt becuase it will reach its final return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY
case, when we look at the behavior past that point, which DOES occur
per the definition of the x86 instruction set, as we have not reached
a "termial" instruction that stops behavior, will see the HHH(DDD)
that DDD called continuing to simulate its input to the point that
this one was defined to stop, and then returns 0 to DDDD and then DDD
returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH
does of the behavior of DDD by its PARTIAL emulation with the ACTUAL
FULL behavior of DDD as defined by the full definition of the x86
insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around
to rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
Which is just your double-talk to try to redefine what halting means.
On 7/12/2024 6:41 PM, Richard Damon wrote:
On 7/12/24 7:19 PM, olcott wrote:
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of
the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the
behavior of DDD, which continues per the definition of the x86
instruction set until the COMPLETE emulation (or direct execution)
reaches a terminal instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have
partial knowledge of the behavior of their DDD, and by returning to
their caller, they establish that behavior for ALL copies of that
HHH, even the one that DDD calls, which shows that DDD will be
halting, even though HHH stopped its observation of the input before
it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a
finite number of steps WILL halt becuase it will reach its final
return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY
case, when we look at the behavior past that point, which DOES occur
per the definition of the x86 instruction set, as we have not
reached a "termial" instruction that stops behavior, will see the
HHH(DDD) that DDD called continuing to simulate its input to the
point that this one was defined to stop, and then returns 0 to DDDD
and then DDD returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH
does of the behavior of DDD by its PARTIAL emulation with the ACTUAL
FULL behavior of DDD as defined by the full definition of the x86
insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around
to rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
Which is just your double-talk to try to redefine what halting means.
You try to cut my airtight proof up in little pieces and fail.
Every rebuttal that you make has disagreeing with the semantics
of the x86 language as its basis.
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of
the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the behavior
of DDD, which continues per the definition of the x86 instruction set
until the COMPLETE emulation (or direct execution) reaches a terminal
instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have partial
knowledge of the behavior of their DDD, and by returning to their
caller, they establish that behavior for ALL copies of that HHH, even
the one that DDD calls, which shows that DDD will be halting, even
though HHH stopped its observation of the input before it gets to that
point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a finite
number of steps WILL halt becuase it will reach its final return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY
case, when we look at the behavior past that point, which DOES occur
per the definition of the x86 instruction set, as we have not reached
a "termial" instruction that stops behavior, will see the HHH(DDD)
that DDD called continuing to simulate its input to the point that
this one was defined to stop, and then returns 0 to DDDD and then DDD
returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH does
of the behavior of DDD by its PARTIAL emulation with the ACTUAL FULL
behavior of DDD as defined by the full definition of the x86
insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around to
rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
Op 13.jul.2024 om 01:19 schreef olcott:
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of
the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the behavior
of DDD, which continues per the definition of the x86 instruction set
until the COMPLETE emulation (or direct execution) reaches a terminal
instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have
partial knowledge of the behavior of their DDD, and by returning to
their caller, they establish that behavior for ALL copies of that
HHH, even the one that DDD calls, which shows that DDD will be
halting, even though HHH stopped its observation of the input before
it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a finite
number of steps WILL halt becuase it will reach its final return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY
case, when we look at the behavior past that point, which DOES occur
per the definition of the x86 instruction set, as we have not reached
a "termial" instruction that stops behavior, will see the HHH(DDD)
that DDD called continuing to simulate its input to the point that
this one was defined to stop, and then returns 0 to DDDD and then DDD
returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH
does of the behavior of DDD by its PARTIAL emulation with the ACTUAL
FULL behavior of DDD as defined by the full definition of the x86
insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around
to rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
This is double talk, because no HHH can possibly exist that simulates
itself correctly.
On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
Op 13.jul.2024 om 01:19 schreef olcott:
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of
the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the
behavior of DDD, which continues per the definition of the x86
instruction set until the COMPLETE emulation (or direct execution)
reaches a terminal instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have
partial knowledge of the behavior of their DDD, and by returning to
their caller, they establish that behavior for ALL copies of that
HHH, even the one that DDD calls, which shows that DDD will be
halting, even though HHH stopped its observation of the input before
it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a
finite number of steps WILL halt becuase it will reach its final
return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY
case, when we look at the behavior past that point, which DOES occur
per the definition of the x86 instruction set, as we have not
reached a "termial" instruction that stops behavior, will see the
HHH(DDD) that DDD called continuing to simulate its input to the
point that this one was defined to stop, and then returns 0 to DDDD
and then DDD returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH
does of the behavior of DDD by its PARTIAL emulation with the ACTUAL
FULL behavior of DDD as defined by the full definition of the x86
insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around
to rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
This is double talk, because no HHH can possibly exist that simulates
itself correctly.
Your definition of correct contradicts the semantics of
the x86 language making it wrong.
Op 13.jul.2024 om 13:39 schreef olcott:
On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
Op 13.jul.2024 om 01:19 schreef olcott:
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of >>>>> the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the
behavior of DDD, which continues per the definition of the x86
instruction set until the COMPLETE emulation (or direct execution)
reaches a terminal instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have
partial knowledge of the behavior of their DDD, and by returning to >>>>> their caller, they establish that behavior for ALL copies of that
HHH, even the one that DDD calls, which shows that DDD will be
halting, even though HHH stopped its observation of the input
before it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a
finite number of steps WILL halt becuase it will reach its final
return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY
case, when we look at the behavior past that point, which DOES
occur per the definition of the x86 instruction set, as we have not >>>>> reached a "termial" instruction that stops behavior, will see the
HHH(DDD) that DDD called continuing to simulate its input to the
point that this one was defined to stop, and then returns 0 to DDDD >>>>> and then DDD returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH
does of the behavior of DDD by its PARTIAL emulation with the
ACTUAL FULL behavior of DDD as defined by the full definition of
the x86 insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around >>>>> to rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
This is double talk, because no HHH can possibly exist that simulates
itself correctly.
Your definition of correct contradicts the semantics of
the x86 language making it wrong.
You have a wrong understanding of the semantics of the x86 language. You think that the x86 language specifies that skipping instructions do not change the behaviour of a program.
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
Op 13.jul.2024 om 13:39 schreef olcott:
On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
Op 13.jul.2024 om 01:19 schreef olcott:
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the >>>>>>> semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior
of the program at the input is a non-aborted one.
And thus HHH that do that know only the first N steps of the
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly. >>>>>>
behavior of DDD, which continues per the definition of the x86
instruction set until the COMPLETE emulation (or direct execution) >>>>>> reaches a terminal instruction.
When we examine the infinite set of every HHH/DDD pair such that: >>>>>>> HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have
partial knowledge of the behavior of their DDD, and by returning
to their caller, they establish that behavior for ALL copies of
that HHH, even the one that DDD calls, which shows that DDD will
be halting, even though HHH stopped its observation of the input
before it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a
finite number of steps WILL halt becuase it will reach its final
return.
The HHH that simulated it for only a finite number of steps, only >>>>>> learned that finite number of steps of the behaivor, and in EVERY >>>>>> case, when we look at the behavior past that point, which DOES
occur per the definition of the x86 instruction set, as we have
not reached a "termial" instruction that stops behavior, will see >>>>>> the HHH(DDD) that DDD called continuing to simulate its input to
the point that this one was defined to stop, and then returns 0 to >>>>>> DDDD and then DDD returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH >>>>>> does of the behavior of DDD by its PARTIAL emulation with the
ACTUAL FULL behavior of DDD as defined by the full definition of
the x86 insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are >>>>>> shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because >>>>>> THEIR DDD will halt. and the HHH that doen't abort never get
around to rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
This is double talk, because no HHH can possibly exist that
simulates itself correctly.
Your definition of correct contradicts the semantics of
the x86 language making it wrong.
You have a wrong understanding of the semantics of the x86 language.
You think that the x86 language specifies that skipping instructions
do not change the behaviour of a program.
You have the wrong understanding of a decider.
All deciders are required to halt.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
Op 13.jul.2024 om 01:19 schreef olcott:
On 7/12/2024 5:56 PM, Richard Damon wrote:
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of
the program at the input is a non-aborted one.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the
behavior of DDD, which continues per the definition of the x86
instruction set until the COMPLETE emulation (or direct execution)
reaches a terminal instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and
aborted its emulation on a non-terminal instrucition only have
partial knowledge of the behavior of their DDD, and by returning to
their caller, they establish that behavior for ALL copies of that
HHH, even the one that DDD calls, which shows that DDD will be
halting, even though HHH stopped its observation of the input before
it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a
finite number of steps WILL halt becuase it will reach its final
return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY
case, when we look at the behavior past that point, which DOES occur
per the definition of the x86 instruction set, as we have not
reached a "termial" instruction that stops behavior, will see the
HHH(DDD) that DDD called continuing to simulate its input to the
point that this one was defined to stop, and then returns 0 to DDDD
and then DDD returning and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH
does of the behavior of DDD by its PARTIAL emulation with the ACTUAL
FULL behavior of DDD as defined by the full definition of the x86
insttuction set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around
to rejecting its DDD as non-halting.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
This is double talk, because no HHH can possibly exist that simulates
itself correctly.
Your definition of correct contradicts the semantics of
the x86 language making it wrong.
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have a wrong understanding of the semantics of the x86 language.
You think that the x86 language specifies that skipping instructions
do not change the behaviour of a program.
You have the wrong understanding of a decider.
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is
because you lie and don't look at the input that this decider actually
has, but look at the input that would have been given to a different
decider to show that one wrong.
On 7/13/24 7:39 AM, olcott wrote:
On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
This is double talk, because no HHH can possibly exist that simulates
itself correctly.
Your definition of correct contradicts the semantics of
the x86 language making it wrong.
No your ideas of the x86 language contradicts the actual sematic of the language.
Where does it ever even imply that a partial emulation correctly
predicts the behavior of the full program?
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have a wrong understanding of the semantics of the x86 language.
You think that the x86 language specifies that skipping instructions
do not change the behaviour of a program.
You have the wrong understanding of a decider.
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the
right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is
because you lie and don't look at the input that this decider actually
has, but look at the input that would have been given to a different
decider to show that one wrong.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
You seem to fail to understand the notion of differing
process contexts. It is a tricky notion for people that
have never done operating system level programming. https://www.geeksforgeeks.org/context-switch-in-operating-system/
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 7:39 AM, olcott wrote:
On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
This is double talk, because no HHH can possibly exist that
simulates itself correctly.
Your definition of correct contradicts the semantics of
the x86 language making it wrong.
No your ideas of the x86 language contradicts the actual sematic of
the language.
Where does it ever even imply that a partial emulation correctly
predicts the behavior of the full program?
You switch from disagreeing with the x86 language to disagreeing
that all deciders must halt.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
On 7/13/24 9:27 AM, olcott wrote:
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have a wrong understanding of the semantics of the x86
language. You think that the x86 language specifies that skipping
instructions do not change the behaviour of a program.
You have the wrong understanding of a decider.
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the
right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is
because you lie and don't look at the input that this decider
actually has, but look at the input that would have been given to a
different decider to show that one wrong.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
But every DDD that calls an HHH that aborts its simulation of a copy of
that DDD and returns is shown to be halting, not non-halting. It is just that HHH can't see that behavior becuase it aborted its simulation.
"DDD" is the program, not the partial emulation of it, so it halts even
if HHHs PARTIAL simulaton of it ddn't reach thatpoint.
You seem to fail to understand the notion of differing
process contexts. It is a tricky notion for people that
have never done operating system level programming.
https://www.geeksforgeeks.org/context-switch-in-operating-system/
Which is something I don't have problems with, since I have written my
own operating systems.
Your problem is you don't seem to understand is that all copies of a
given deterministic program act the same
On 7/13/2024 9:14 AM, Richard Damon wrote:
On 7/13/24 9:27 AM, olcott wrote:
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have a wrong understanding of the semantics of the x86
language. You think that the x86 language specifies that skipping >>>>>> instructions do not change the behaviour of a program.
You have the wrong understanding of a decider.
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the
right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is
because you lie and don't look at the input that this decider
actually has, but look at the input that would have been given to a
different decider to show that one wrong.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
But every DDD that calls an HHH that aborts its simulation of a copy
of that DDD and returns is shown to be halting, not non-halting. It is
just that HHH can't see that behavior becuase it aborted its simulation.
"DDD" is the program, not the partial emulation of it, so it halts
even if HHHs PARTIAL simulaton of it ddn't reach thatpoint.
You seem to fail to understand the notion of differing
process contexts. It is a tricky notion for people that
have never done operating system level programming.
https://www.geeksforgeeks.org/context-switch-in-operating-system/
Which is something I don't have problems with, since I have written my
own operating systems.
Your problem is you don't seem to understand is that all copies of a
given deterministic program act the same
By this same reasoning when you are hungry and eat until
you are full you are still hungry because you are still
yourself.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
On 7/13/24 10:35 AM, olcott wrote:
On 7/13/2024 9:14 AM, Richard Damon wrote:
On 7/13/24 9:27 AM, olcott wrote:
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have a wrong understanding of the semantics of the x86
language. You think that the x86 language specifies that skipping >>>>>>> instructions do not change the behaviour of a program.
You have the wrong understanding of a decider.
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the
right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is
because you lie and don't look at the input that this decider
actually has, but look at the input that would have been given to a >>>>> different decider to show that one wrong.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
But every DDD that calls an HHH that aborts its simulation of a copy
of that DDD and returns is shown to be halting, not non-halting. It
is just that HHH can't see that behavior becuase it aborted its
simulation.
"DDD" is the program, not the partial emulation of it, so it halts
even if HHHs PARTIAL simulaton of it ddn't reach thatpoint.
You seem to fail to understand the notion of differing
process contexts. It is a tricky notion for people that
have never done operating system level programming.
https://www.geeksforgeeks.org/context-switch-in-operating-system/
Which is something I don't have problems with, since I have written
my own operating systems.
Your problem is you don't seem to understand is that all copies of a
given deterministic program act the same
By this same reasoning when you are hungry and eat until
you are full you are still hungry because you are still
yourself.
Not at all, I am not a deterministic entity like HHH and DDD.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
On 7/13/2024 9:48 AM, Richard Damon wrote:
On 7/13/24 10:35 AM, olcott wrote:
On 7/13/2024 9:14 AM, Richard Damon wrote:
On 7/13/24 9:27 AM, olcott wrote:
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have the wrong understanding of a decider.
You have a wrong understanding of the semantics of the x86
language. You think that the x86 language specifies that
skipping instructions do not change the behaviour of a program. >>>>>>>
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the >>>>>> right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is >>>>>> because you lie and don't look at the input that this decider
actually has, but look at the input that would have been given to >>>>>> a different decider to show that one wrong.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
But every DDD that calls an HHH that aborts its simulation of a copy
of that DDD and returns is shown to be halting, not non-halting. It
is just that HHH can't see that behavior becuase it aborted its
simulation.
"DDD" is the program, not the partial emulation of it, so it halts
even if HHHs PARTIAL simulaton of it ddn't reach thatpoint.
You seem to fail to understand the notion of differing
process contexts. It is a tricky notion for people that
have never done operating system level programming.
https://www.geeksforgeeks.org/context-switch-in-operating-system/
Which is something I don't have problems with, since I have written
my own operating systems.
Your problem is you don't seem to understand is that all copies of a
given deterministic program act the same
By this same reasoning when you are hungry and eat until
you are full you are still hungry because you are still
yourself.
Not at all, I am not a deterministic entity like HHH and DDD.
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Try and use this free will to make a square circle.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
On 7/13/24 11:15 AM, olcott wrote:
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Just shows that though I have free will, I am also in a Universe with a
lot of determinism.
Try and use this free will to make a square circle.
Nope, just shows you don't know what you are talking about and need to switch to Red Herring because you lost the argument.
Face it, all you have proved is that you are nothing but a pathetic
ignorant pathological lying idiot.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
Right, but the program it represents, and the question is about IS.
On 7/13/2024 10:25 AM, Richard Damon wrote:
On 7/13/24 11:15 AM, olcott wrote:
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Just shows that though I have free will, I am also in a Universe with
a lot of determinism.
Try and use this free will to make a square circle.
Nope, just shows you don't know what you are talking about and need to
switch to Red Herring because you lost the argument.
Face it, all you have proved is that you are nothing but a pathetic
ignorant pathological lying idiot.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
Right, but the program it represents, and the question is about IS.
HHH cannot be correctly required to report on the behavior
of an external process that it has no access to.
As soon as HHH correctly determines that it must abort the
simulation of its input to prevent its own infinite execution
HHH is necessarily correct to reject this finite string as
specifying non-halting behavior.
On 7/13/24 11:34 AM, olcott wrote:
On 7/13/2024 10:25 AM, Richard Damon wrote:
On 7/13/24 11:15 AM, olcott wrote:
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Just shows that though I have free will, I am also in a Universe with
a lot of determinism.
Try and use this free will to make a square circle.
Nope, just shows you don't know what you are talking about and need
to switch to Red Herring because you lost the argument.
Face it, all you have proved is that you are nothing but a pathetic
ignorant pathological lying idiot.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
Right, but the program it represents, and the question is about IS.
HHH cannot be correctly required to report on the behavior
of an external process that it has no access to.
But it has access to the complete representation of it.
On 7/13/2024 11:05 AM, Richard Damon wrote:
On 7/13/24 11:34 AM, olcott wrote:
On 7/13/2024 10:25 AM, Richard Damon wrote:
On 7/13/24 11:15 AM, olcott wrote:
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Just shows that though I have free will, I am also in a Universe
with a lot of determinism.
Try and use this free will to make a square circle.
Nope, just shows you don't know what you are talking about and need
to switch to Red Herring because you lost the argument.
Face it, all you have proved is that you are nothing but a pathetic
ignorant pathological lying idiot.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
Right, but the program it represents, and the question is about IS.
HHH cannot be correctly required to report on the behavior
of an external process that it has no access to.
But it has access to the complete representation of it.
In other words you are still hungry AFTER you filled
yourself with food BECAUSE you are the same person
thus the change in process state DOES NOT MATTER.
On 7/13/24 12:19 PM, olcott wrote:
On 7/13/2024 11:05 AM, Richard Damon wrote:
On 7/13/24 11:34 AM, olcott wrote:
On 7/13/2024 10:25 AM, Richard Damon wrote:
On 7/13/24 11:15 AM, olcott wrote:
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Just shows that though I have free will, I am also in a Universe
with a lot of determinism.
Try and use this free will to make a square circle.
Nope, just shows you don't know what you are talking about and need >>>>> to switch to Red Herring because you lost the argument.
Face it, all you have proved is that you are nothing but a pathetic >>>>> ignorant pathological lying idiot.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
Right, but the program it represents, and the question is about IS.
HHH cannot be correctly required to report on the behavior
of an external process that it has no access to.
But it has access to the complete representation of it.
In other words you are still hungry AFTER you filled
yourself with food BECAUSE you are the same person
thus the change in process state DOES NOT MATTER.
Maybe you need to stop eating so much Herring with Red Sauce, and focus
on some of the errors pointed out in your logic rather than just
ignoring them, which, in effect, just admitss that you have no idea how
to get out of your lies.
On 7/13/2024 11:31 AM, Richard Damon wrote:
On 7/13/24 12:19 PM, olcott wrote:
On 7/13/2024 11:05 AM, Richard Damon wrote:
On 7/13/24 11:34 AM, olcott wrote:
On 7/13/2024 10:25 AM, Richard Damon wrote:
On 7/13/24 11:15 AM, olcott wrote:
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Just shows that though I have free will, I am also in a Universe
with a lot of determinism.
Try and use this free will to make a square circle.
Nope, just shows you don't know what you are talking about and
need to switch to Red Herring because you lost the argument.
Face it, all you have proved is that you are nothing but a
pathetic ignorant pathological lying idiot.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
Right, but the program it represents, and the question is about IS. >>>>>>
HHH cannot be correctly required to report on the behavior
of an external process that it has no access to.
But it has access to the complete representation of it.
In other words you are still hungry AFTER you filled
yourself with food BECAUSE you are the same person
thus the change in process state DOES NOT MATTER.
Maybe you need to stop eating so much Herring with Red Sauce, and
focus on some of the errors pointed out in your logic rather than just
ignoring them, which, in effect, just admitss that you have no idea
how to get out of your lies.
You continue to stupidly insist that DDD specifies
the same behavior before its simulation has been
aborted than after it simulation has been aborted.
Now you stupidly insist that this is not analogous
to being hungry before you have eaten and not being
hungry after you have eaten.
That you call me a liar will send you to Hell.
I don't want that.
On 7/13/24 12:43 PM, olcott wrote:
On 7/13/2024 11:31 AM, Richard Damon wrote:
On 7/13/24 12:19 PM, olcott wrote:
On 7/13/2024 11:05 AM, Richard Damon wrote:
On 7/13/24 11:34 AM, olcott wrote:
On 7/13/2024 10:25 AM, Richard Damon wrote:
On 7/13/24 11:15 AM, olcott wrote:
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Just shows that though I have free will, I am also in a Universe >>>>>>> with a lot of determinism.
Try and use this free will to make a square circle.
Nope, just shows you don't know what you are talking about and
need to switch to Red Herring because you lost the argument.
Face it, all you have proved is that you are nothing but a
pathetic ignorant pathological lying idiot.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
Right, but the program it represents, and the question is about IS. >>>>>>>
HHH cannot be correctly required to report on the behavior
of an external process that it has no access to.
But it has access to the complete representation of it.
In other words you are still hungry AFTER you filled
yourself with food BECAUSE you are the same person
thus the change in process state DOES NOT MATTER.
Maybe you need to stop eating so much Herring with Red Sauce, and
focus on some of the errors pointed out in your logic rather than
just ignoring them, which, in effect, just admitss that you have no
idea how to get out of your lies.
You continue to stupidly insist that DDD specifies
the same behavior before its simulation has been
aborted than after it simulation has been aborted.
And you think that HHH partial observation of the some of the behavior
of DDD affects it.
On 7/13/2024 12:20 PM, Richard Damon wrote:
On 7/13/24 12:43 PM, olcott wrote:
On 7/13/2024 11:31 AM, Richard Damon wrote:
On 7/13/24 12:19 PM, olcott wrote:
On 7/13/2024 11:05 AM, Richard Damon wrote:
On 7/13/24 11:34 AM, olcott wrote:
On 7/13/2024 10:25 AM, Richard Damon wrote:
On 7/13/24 11:15 AM, olcott wrote:
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Just shows that though I have free will, I am also in a Universe >>>>>>>> with a lot of determinism.
Try and use this free will to make a square circle.
Nope, just shows you don't know what you are talking about and >>>>>>>> need to switch to Red Herring because you lost the argument.
Face it, all you have proved is that you are nothing but a
pathetic ignorant pathological lying idiot.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
Right, but the program it represents, and the question is about IS. >>>>>>>>
HHH cannot be correctly required to report on the behavior
of an external process that it has no access to.
But it has access to the complete representation of it.
In other words you are still hungry AFTER you filled
yourself with food BECAUSE you are the same person
thus the change in process state DOES NOT MATTER.
Maybe you need to stop eating so much Herring with Red Sauce, and
focus on some of the errors pointed out in your logic rather than
just ignoring them, which, in effect, just admitss that you have no
idea how to get out of your lies.
You continue to stupidly insist that DDD specifies
the same behavior before its simulation has been
aborted than after it simulation has been aborted.
And you think that HHH partial observation of the some of the behavior
of DDD affects it.
No stupid you know that I didn't say anything like that.
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