olcott <NoOne@NoWhere.com> writes:
On 3/16/2022 8:16 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/12/2022 8:55 PM, Ben Bacarisse wrote:
André G. Isaak <agisaak@gm.invalid> writes:
I once tried to get a direct answer to this question. I asked 12 times >>>>> in consecutive posts but never got one.On 3/12/2022 5:57 PM, André G. Isaak wrote:
So what string, according to you, encodes the computation Ĥ applied >>>>>>>> to ⟨Ĥ⟩? If these two "different" computations don't have separate >>>>>>>> encodings as strings then they are not, in fact, different
computations at all.
No Comment?
I know you've been asked this question before and have consistently >>>>>> ignored it. According to a recent post of yours that constitutes
justification for a repetitive all-caps temper tantrum!
Later, on the related question of whether ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
computation I got this dazzling display of equivocation:
"When it is construed as input to H then ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
computation.
When it is construed as input to Ĥ.qx then ⟨Ĥ⟩ ⟨Ĥ⟩ DOES NOT encode a
halting computation."
Bear in mind that at time, PO's machines were magic: two identical state >>>>> transition functions could entail transitions to different states when >>>>> presented with identical inputs. He has since backed off from some of >>>>> these remarks, but it never exactly clear which previous claims he would >>>>> now accept were wrong.
None-the-less...
You mean you won't comment on the above but would rather present new
junk about BASIC. Oh well... I can't stop you.
None-the-less none of what you have ever said shows that I am
incorrect.
Ah! So from your point of view I did not point out an error in the post
you replied to. That means you /still/ think that:
"When it is construed as input to H then ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
computation. When it is construed as input to Ĥ.qx then ⟨Ĥ⟩ ⟨Ĥ⟩ DOES
NOT encode a halting computation."
That's useful to know. Either your strings or your TMs are still
magic. Sadly, that means you have nothing to say about the halting
problem for Turing machines. You need to fix that before anyone can
take you seriously.
On 3/16/22 9:51 PM, olcott wrote:embedded_H has all of the functionality of a UTM and is able to
On 3/16/2022 8:36 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/16/2022 8:16 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/12/2022 8:55 PM, Ben Bacarisse wrote:
André G. Isaak <agisaak@gm.invalid> writes:
I once tried to get a direct answer to this question. I asked 12 >>>>>>> timesOn 3/12/2022 5:57 PM, André G. Isaak wrote:
So what string, according to you, encodes the computation Ĥ >>>>>>>>>> applied
to ⟨Ĥ⟩? If these two "different" computations don't have separate
encodings as strings then they are not, in fact, different >>>>>>>>>> computations at all.
No Comment?
I know you've been asked this question before and have consistently >>>>>>>> ignored it. According to a recent post of yours that constitutes >>>>>>>> justification for a repetitive all-caps temper tantrum!
in consecutive posts but never got one.
Later, on the related question of whether ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
computation I got this dazzling display of equivocation:
"When it is construed as input to H then ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a
halting
computation.
When it is construed as input to Ĥ.qx then ⟨Ĥ⟩ ⟨Ĥ⟩ DOES NOT
encode a
halting computation."
Bear in mind that at time, PO's machines were magic: two
identical state
transition functions could entail transitions to different states >>>>>>> when
presented with identical inputs. He has since backed off from
some of
these remarks, but it never exactly clear which previous claims
he would
now accept were wrong.
None-the-less...
You mean you won't comment on the above but would rather present new >>>>> junk about BASIC. Oh well... I can't stop you.
None-the-less none of what you have ever said shows that I am
incorrect.
Ah! So from your point of view I did not point out an error in the post >>> you replied to. That means you /still/ think that:
"When it is construed as input to H then ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
computation. When it is construed as input to Ĥ.qx then ⟨Ĥ⟩ ⟨Ĥ⟩ DOES
NOT encode a halting computation."
THIS IS THE ONLY POINT THAT MATTERS
THIS IS THE ONLY POINT THAT MATTERS
THIS IS THE ONLY POINT THAT MATTERS
THIS IS THE ONLY POINT THAT MATTERS
That the simulated input: ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never reach the
final state of this simulated input in any finite number of steps of
correct simulation by embedded_H conclusively proves that a mapping
from this input to the reject state of embedded_H is correct.
Except that 'correct simulation by embedded_H' only has meaning if
embedded_H is actually
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