Re: H(D,D) cannot even be asked about the behavior of D(D) V3 ---IGNORING ALL OTHER REPLIES

On 2024-06-16 01:42:29 +0000, olcott said:

On 6/15/2024 8:19 PM, Richard Damon wrote:

On 6/15/24 8:48 PM, olcott wrote:

On 6/15/2024 7:13 PM, Richard Damon wrote:

On 6/15/24 8:05 PM, olcott wrote:

On 6/15/2024 6:37 PM, Richard Damon wrote:

On 6/15/24 7:30 PM, olcott wrote:

On 6/15/2024 6:01 PM, Richard Damon wrote:

On 6/15/24 5:56 PM, olcott wrote:

On 6/15/2024 11:33 AM, Richard Damon wrote:

On 6/15/24 12:22 PM, olcott wrote:

On 6/13/2024 8:24 PM, Richard Damon wrote:
 > On 6/13/24 11:32 AM, olcott wrote:
 >>
 >> It is contingent upon you to show the exact steps of how H computes
 >> the mapping from the x86 machine language finite string input to
 >> H(D,D) using the finite string transformation rules specified by
 >> the semantics of the x86 programming language that reaches the
 >> behavior of the directly executed D(D)
 >>
 >
 > Why? I don't claim it can.
 The first six steps of this mapping are when instructions
at the machine address range of [00000cfc] to [00000d06]
are simulated/executed.
 After that the behavior of D correctly simulated by H diverges
from the behavior of D(D) because the call to H(D,D) by D
correctly simulated by H cannot possibly return to D.

 Nope, the steps of D correctly simulated by H will EXACTLY match the steps of D directly executed, until H just gives up and guesses.
 

 When we can see that D correctly simulated by H cannot possibly
reach its simulated final state at machine address [00000d1d]
after one recursive simulation and the same applies for 2,3,...N
recursive simulations then we can abort the simulated input and
correctly report that D correctly simulated by H DOES NOT HALT.

 Nope. Because an aborted simulation doesn't say anything about Halting,
 

 It is the mathematical induction that says this.
 

WHAT "Mathematical Induction"?
 

 A proof by induction consists of two cases. The first, the base
case, proves the statement for n = 0 without assuming any knowledge
of other cases. The second case, the induction step, proves that
if the statement holds for any given case n = k then it must also
hold for the next case n = k + 1 These two steps establish that the
statement holds for every natural number n.
https://en.wikipedia.org/wiki/Mathematical_induction

 Ok, so you can parrot to words.
 

 It is true that after one recursive simulation of D correctly
simulated by H that D does not reach its simulated final state
at machine address [00000d1d].

 Which means you consider that D has been bound to that first H, so you have instruciton to simulate in the call H.
 

 *We directly see this is true for every N thus no assumption needed*
It is true that after N recursive simulations of D correctly
simulated by H that D does not reach its simulated final state
at machine address [00000d1d].

 Nope, because to do the first step, you had to bind the definition of the first H to D, and thus can not change it.

 So infinite sets are permanently beyond your grasp.
The above D simulated by any H has the same property
of never reaching its own simulated machine address
at [00000d1d].
 What I mistook for dishonestly is simply a lack
of comprehension.
 

  But it isn't an infinite set.
 

 Sure it is you are just clueless.
I mistook your ignorance for deception.
 

We don't ask an infinite set a question, or give a decider an infinite set of inputs.
 

 Yes we do and this is simply over your head.
 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 The second ⊢* wildcard specifies this infinite set.

As you should already know, ⊢* as used by Linz is not a wildcard.
It is a repeated application of ⊢ without showing intermediate steps.
--
Mikko