On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:
That is not true. The pattern exists for at least any finite number ofHere is what it actually does:
These steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩
⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩
⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
⟨Ĥ5⟩...
And if that is what it actually does, then H NEVER aborts its
simulation and thus never give an answer.
When embedded_H correctly matches the above infinite sequence this
conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).
Excepts as previously said, that pattern only exists if H never aborts.
steps where it can be recognized. The three iterations shown above are
plenty enough for it to be recogized.
But if it only exists for a finite number of steps (till it is
recognized)
On 2/3/22 10:10 PM, olcott wrote:
On 2/3/2022 5:50 AM, Richard Damon wrote:
On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:
That is not true. The pattern exists for at least any finite numberHere is what it actually does:
These steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates
⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates
⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates
⟨Ĥ4⟩ ⟨Ĥ5⟩...
And if that is what it actually does, then H NEVER aborts its
simulation and thus never give an answer.
When embedded_H correctly matches the above infinite sequence this >>>>>> conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM). >>>>>>
Excepts as previously said, that pattern only exists if H never
aborts.
of steps where it can be recognized. The three iterations shown
above are plenty enough for it to be recogized.
But if it only exists for a finite number of steps (till it is
recognized)
We are discussing the point in the execution of embedded_H where it
has just correctly matched an infinite behavior pattern while it was
doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩
No, you are CLAIMING (incorrectly) that it has made that determination.
On 2/3/22 10:56 PM, olcott wrote:So in other words the concept of logical necessity is so far over your
On 2/3/2022 9:40 PM, Richard Damon wrote:
On 2/3/22 10:10 PM, olcott wrote:
On 2/3/2022 5:50 AM, Richard Damon wrote:
On 2/3/22 12:24 AM, olcott wrote:
On 2/2/2022 11:09 PM, Richard Damon wrote:
On 2/2/22 11:50 PM, olcott wrote:
On 2/2/2022 10:42 PM, Richard Damon wrote:
On 2/2/22 10:50 PM, olcott wrote:
That is not true. The pattern exists for at least any finiteHere is what it actually does:
These steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates
⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates
⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates
⟨Ĥ4⟩ ⟨Ĥ5⟩...
And if that is what it actually does, then H NEVER aborts its >>>>>>>>> simulation and thus never give an answer.
When embedded_H correctly matches the above infinite sequence
this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩
cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any >>>>>>>> UTM).
Excepts as previously said, that pattern only exists if H never
aborts.
number of steps where it can be recognized. The three iterations
shown above are plenty enough for it to be recogized.
But if it only exists for a finite number of steps (till it is
recognized)
We are discussing the point in the execution of embedded_H where it
has just correctly matched an infinite behavior pattern while it was
doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>
No, you are CLAIMING (incorrectly) that it has made that determination.
THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
As soon as an infinite behavior pattern is correctly recognized in a
finite number of steps then it is definitely correct for embedded_H to
transition to Ĥ.qn.
Except that such a pattern in H^ is a Fairy Dust Powered Unicorn,
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