Liste des Groupes | Revenir à theory |
Am Fri, 07 Jun 2024 14:31:10 -0500 schrieb olcott:>On 6/7/2024 1:57 PM, wij wrote:On Fri, 2024-06-07 at 13:41 -0500, olcott wrote:When we can show that even in the halting problem HH is only required toOn 6/7/2024 1:24 PM, Richard Damon wrote:The Halting Problem asks for a program H (precisely a TM) that:On 6/7/24 2:02 PM, olcott wrote:>On 6/7/2024 12:50 PM, Alan Mackenzie wrote:And thus you admit that HH is not a Halt Decider,In comp.theory olcott <polcott333@gmail.com> wrote:>
Anyone claiming that HH should report on the behavior of the
directly executed DD(DD) is requiring a violation of the above
definition of correct simulation.
>
IF H(D,D)==1, THEN D(D) will return.
ELSE If H(D,D)==0, THEN D(D) will never return.
ELSE HP is undecidable
>
report on the behavior of DD correctly simulated by HH these dishonest
people merely use that as another deflection point for their dishonesty.
The way around this that just worked is to stay diligently focused one
one single point until the dishonest people finally admit that they have
simply ignored all the proofs for three solid years.
"only" It must report on the behaviour of DD, which must be the same whenThe most persistent false assumption that cannot possibly
simulated. It can't simulate something different and say "look! My result
simulating this is right, because it is my result!".
--The fact that the execution trace of P derived by the executed H and theDoes the called H also match?
simulated H exactly matches the machine code of P proves that each
instruction of P was simulated correctly and in the correct order this
conclusively proves that P is correctly simulated by both of these
instances of H.
Les messages affichés proviennent d'usenet.