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On 3/10/2024 8:33 AM, Mikko wrote:Nice to see that you don't disagree about verified facts.On 2024-03-10 01:29:39 +0000, olcott said:Execution trace of Ĥ applied to ⟨Ĥ⟩
On 3/9/2024 7:24 PM, immibis wrote:"Ĥ.q0 ⟨Ĥ⟩ does not halt // Ĥ applied to ⟨Ĥ⟩ halts" is not aOn 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: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.On 10/03/24 01:22, olcott wrote:It is objectively true that Ĥ.H can get stuck in recursiveOn 3/9/2024 5:57 PM, immibis wrote: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.On 10/03/24 00:26, olcott wrote:The above is true no matter what criteria that is usedOn 3/9/2024 5:10 PM, immibis wrote:You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.On 9/03/24 23:22, olcott wrote:*Only because Ĥ.H is embedded within Ĥ and H is not*On 3/9/2024 3:50 PM, immibis wrote:Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.On 9/03/24 22:34, olcott wrote:Simulating halt deciders must make sure that they themselvesWhat criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows whatĤ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.
wrong answer to provide?
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⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
as long as H is a simulating halt decider.
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.
*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
verified fact.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt"Ĥ.q0 ⟨Ĥ⟩ halts // Ĥ applied to ⟨Ĥ⟩ does not halt" is not a
verified fact.
(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
The earliest point when Turing machine Ĥ can detect the repeating
state of its input is when Ĥ reaches (c) a second time where its
input would begin simulating a copy of itself with a copy of its
input. It could detect this one execution trace earlier
[ when its input first reaches (c) ] if Ĥ was an Olcott machine.
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