olcott <NoOne@NoWhere.com> writes:
As soon as we verify that the correct pure simulation of the input to
H(P,0) never reaches the last address of P at c50 we know that the
input to H(P,P) never halts thus the correctly halt status of
H(P,P)==0;
But P(P) halts, so H(P,P) == 0 is the wrong result.
An H this is wrong
about this key case is not interesting, unlike your H from Dec 2018:
"Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz specs
does not exist. I now have a fully encoded pair of Turing Machines H / Ĥ
proving them wrong."[1]
Three years of progressively walking back that claim and you are left
with some C function H that is wrong about H(Ĥ, Ĥ). No one thinks such
an H does not exist. It's trivially obvious that an H that does /not/
meet Linz's spec (for the (Ĥ, Ĥ) case) exists. They are ten-a-penny.
[1] Message-ID: <JbydneYGrrpProjBnZ2dnUU7-X3NnZ2d@giganews.com>
On Saturday, 6 November 2021 at 12:21:49 UTC, olcott wrote:
On 11/5/2021 6:49 PM, Ben Bacarisse wrote:The first seven steps are indeed simulated. But the next seven steps are simulated by the copy of H that is called from P.
olcott <No...@NoWhere.com> writes:When the correct pure simulation of the actual input to the halt decider
As soon as we verify that the correct pure simulation of the input to
H(P,0) never reaches the last address of P at c50 we know that the
input to H(P,P) never halts thus the correctly halt status of
H(P,P)==0;
But P(P) halts, so H(P,P) == 0 is the wrong result.
never halts then the halt decider is necessarily correct when it reports
that its input never halts.
Disagreeing with a logical tautology is a break from reality.
A correct pure simulation is not verified on the basis of any mere
assumptions. As long as the first seven x86 instructions of the source
code of P are executed in order and the seventh instruction causes this
cycle to repeat then we know that H(P,P) performed a correct pure
simulation of its input for the entire 14 steps of this simulation.
So they shouldn't appear
in the instruction trace. If we simulate "push ebp" we don't actually push ebp to the stack. We create a virtual stack in memory and push a value to
it.
Sysop: | Keyop |
---|---|
Location: | Huddersfield, West Yorkshire, UK |
Users: | 366 |
Nodes: | 16 (2 / 14) |
Uptime: | 17:07:40 |
Calls: | 7,812 |
Files: | 12,927 |
Messages: | 5,766,217 |