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On 3/9/2024 7:24 PM, immibis wrote:That's not a verified fact, that's just something you want to be true.On 10/03/24 01:30, olcott wrote:*In other words you are denying these verified facts*On 3/9/2024 6:24 PM, immibis wrote:>On 10/03/24 01:22, olcott wrote:>On 3/9/2024 5:57 PM, immibis wrote:>On 10/03/24 00:26, olcott wrote:>On 3/9/2024 5:10 PM, immibis wrote:>On 9/03/24 23:22, olcott wrote:>On 3/9/2024 3:50 PM, immibis wrote:>On 9/03/24 22:34, olcott wrote:>>>
What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows what
wrong answer to provide?
>
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.
Simulating halt deciders must make sure that they themselves
do not get stuck in infinite execution. This means that they
must abort every simulation that cannot possibly otherwise halt.
>
This requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts
its simulation.
>
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H ⟨Ĥ⟩ ⟨Ĥ⟩
does not simulate itself in recursive simulation.
>
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.
>
*Only because Ĥ.H is embedded within Ĥ and H is not*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
>
You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.
The above is true no matter what criteria that is used
as long as H is a simulating halt decider.
>
Objective criteria cannot vary based on who the subject is. They are objective. The answer to different people is the same answer if the criteria are objective.
It is objectively true that Ĥ.H can get stuck in recursive
simulation because Ĥ copies its input thus never runs
out of params.
>
It is objectively true that Ĥ cannot possibly get stuck
in recursive because H does not copy its input thus runs
out of params.
>
Wrong. Dead wrong. Stupidly wrong. So wrong that a dead monkey could do better. Write the Olcott machine (not x86utm) code for Ĥ and I would show you.
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Execution trace of Ĥ applied to ⟨Ĥ⟩Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ BECAUSE IT IS PRECISELY IDENTICAL TO STEPS B AND C:
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
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