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

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.
If we simulate THAT input for one more step, it will see the FIRST H decide to abort and return.
If you don't bind the first H, yo can't do the first step.

 _D()
[00000cfc](01) 55          push ebp
[00000cfd](02) 8bec        mov ebp,esp
[00000cff](03) 8b4508      mov eax,[ebp+08]
[00000d02](01) 50          push eax       ; push D
[00000d03](03) 8b4d08      mov ecx,[ebp+08]
[00000d06](01) 51          push ecx       ; push D
[00000d07](05) e800feffff  call 00000b0c  ; call H
[00000d0c](03) 83c408      add esp,+08
[00000d0f](02) 85c0        test eax,eax
[00000d11](02) 7404        jz 00000d17
[00000d13](02) 33c0        xor eax,eax
[00000d15](02) eb05        jmp 00000d1c
[00000d17](05) b801000000  mov eax,00000001
[00000d1c](01) 5d          pop ebp
[00000d1d](01) c3          ret
Size in bytes:(0034) [00000d1d]
 

You haven't shown the required pieces for an inductive proof.
>
I doubt you even know what you need to do, let alone be able to do it.