Re: Two dozen people were simply wrong --- Try to prove otherwise

On 2024-06-03 13:16:12 +0000, olcott said:

On 6/3/2024 5:03 AM, Mikko wrote:

On 2024-05-31 15:13:02 +0000, olcott said:
 

On 5/31/2024 3:30 AM, Mikko wrote:

On 2024-05-31 01:54:52 +0000, olcott said:
 

On 5/30/2024 8:37 PM, Richard Damon wrote:

On 5/30/24 9:31 AM, olcott wrote:

On 5/30/2024 2:40 AM, Mikko wrote:

On 2024-05-30 01:15:21 +0000, olcott said:

 

x <is> a finite string Turing machine description that SPECIFIES behavior. The term: "representing" is inaccurate.

 No, x is a description of the Turing machine that specifies the behaviour
that H is required to report.

 That is what I said.

 Note, the string doesn't DIRECTLY specify behavior, but only indirectly as a description/representation of the Turing Mach
 

 The string directly SPECIFIES behavior to a UTM or to
any TM based on a UTM.

 An UTM interpretes the string as a specification of behaviour

 YES, exactly !!!
 

and another Turing machine may interprete likewise. But in a
different context the interpretation is different.
 

 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 When embedded_H <is> a UTM or <is> a halting computation based on a
UTM then the ⟨Ĥ⟩ ⟨Ĥ⟩ input to embedded_H SPECIFIES that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
simulated by embedded_H cannot possibly reach its own simulated final
state at ⟨Ĥ.qn⟩.

 There is no requirement to follow the pecifications by embedded_H.

  In other words embedded_H can simply play bingo and never halt and still correctly decide the halt status of its input?

Not of the input text but of the machine and input that the text
describes. The input to H contains enough information that the
machine can be constructed and run with the input. If the prediction
by H differs from the actual execution of the machine the prediction
is wrong and H is not a halt decider.
--
Mikko