Re: H(D,D) cannot even be asked about the behavior of D(D) V2

On 6/16/2024 9:16 AM, joes wrote:

Am Sun, 16 Jun 2024 07:44:41 -0500 schrieb olcott:

On 6/16/2024 2:50 AM, Mikko wrote:

On 2024-06-15 13:14:57 +0000, olcott said:

On 6/15/2024 7:19 AM, Mikko wrote:

On 2024-06-15 03:07:14 +0000, olcott said:

On 6/13/2024 8:24 PM, Richard Damon wrote:
  > On 6/13/24 11:32 AM, olcott wrote:

 

Whenever a decider is run it answers the question it is made to answer.

Not necessarily. Just because everyone falsely assumes that D correctly
simulated by H must have the same behavior as the directly executed D(D)
does not make this false assumption true.

You still need to explain how you can call a simulation that differs from
the behaviour of its input "correct".
 

I have proven it many times and this proof is simply over
everyone's heads. When I ask what your C programming skill
level is, this *is not* a rhetorical question.
00   typedef void (*ptr)(); // pointer to void function
01
02   int H0(ptr P);
03
04   void DDD()
05   {
06     H0(DDD);
07     return;
08   }
09
10   int main()
11   {
12     H0(DDD);
13   }
Line 12 main()
   invokes H0(DDD); that simulates DDD()
*REPEAT UNTIL outer H0 aborts*
   Line 06 simulated DDD()
   invokes simulated H0(DDD); that simulates DDD()
DDD correctly simulated by H0 never reaches its own "return"
instruction and halts.

H is only asked about the behavior of D simulated by H and is
not asked about the behavior of the directly executed D(D). >>

If H is a simulator, it must simulate the execution of D(D).

I have proven this to be a false assumption and people
maintain this false assumption entirely on the basis
that my proof is over their heads.

H does not compute the answer to "What does H say about its input?",
since it could answer anything then.
It makes no sense to call a wrong answer the correct answer to a different
question.
 

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer