XPost: comp.theory, sci.logic   
   From: richard@damon-family.org   
      
   On 3/13/24 10:20 PM, olcott wrote:   
   > On 3/13/2024 1:16 PM, Ross Finlayson wrote:   
   >> On 03/12/2024 09:00 PM, olcott wrote:   
   >>> On 3/12/2024 10:49 PM, Ross Finlayson wrote:   
   >>>> On 03/12/2024 08:23 PM, Ross Finlayson wrote:   
   >>>>> On 03/12/2024 07:52 PM, olcott wrote:   
   >>>>>> On 3/12/2024 9:28 PM, Richard Damon wrote:   
   >>>>>>> On 3/12/24 4:31 PM, olcott wrote:   
   >>>>>>>> On 3/12/2024 6:11 PM, Richard Damon wrote:   
   >>>>>>>>> On 3/12/24 3:53 PM, olcott wrote:   
   >>>>>>>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:   
   >>>>>>>>>>> On 3/12/24 2:34 PM, olcott wrote:   
   >>>>>>>>>>>> On 3/12/2024 4:23 PM, Richard Damon wrote:   
   >>>>>>>>>>>>> On 3/12/24 1:11 PM, olcott wrote:   
   >>>>>>>>>>>>>> On 3/12/2024 2:40 PM, Richard Damon wrote:   
   >>>>>>>>>>>>>>> On 3/12/24 12:02 PM, olcott wrote:   
   >>>>>>>>>>>>>>>> On 3/12/2024 1:31 PM, immibis wrote:   
   >>>>>>>>>>>>>>>>> On 12/03/24 19:12, olcott wrote:   
   >>>>>>>>>>>>>>>>>> ∀ H ∈ Turing_Machine_Deciders   
   >>>>>>>>>>>>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |   
   >>>>>>>>>>>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)   
   >>>>>>>>>>>>>>>>>>   
   >>>>>>>>>>>>>>>>>> There is some input TMD to every H such that   
   >>>>>>>>>>>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)   
   >>>>>>>>>>>>>>>>>   
   >>>>>>>>>>>>>>>>> And it can be a different TMD to each H.   
   >>>>>>>>>>>>>>>>>   
   >>>>>>>>>>>>>>>>>> When we disallow decider/input pairs that are incorrect   
   >>>>>>>>>>>>>>>>>> questions where both YES and NO are the wrong answer   
   >>>>>>>>>>>>>>>>>   
   >>>>>>>>>>>>>>>>> Once we understand that either YES or NO is the right   
   >>>>>>>>>>>>>>>>> answer, the whole rebuttal is tossed out as invalid and   
   >>>>>>>>>>>>>>>>> incorrect.   
   >>>>>>>>>>>>>>>>>   
   >>>>>>>>>>>>>>>>   
   >>>>>>>>>>>>>>>> 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   
   >>>>>>>>>>>>>>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩   
   ⟨Ĥ⟩   
   >>>>>>>>>>>>>>>   
   >>>>>>>>>>>>>>> No, because a given H will only go to one of the answers.   
   >>>>>>>>>>>>>>> THAT   
   >>>>>>>>>>>>>>> will be wrong, and the other one right.   
   >>>>>>>>>>>>>>>   
   >>>>>>>>>>>>>>   
   >>>>>>>>>>>>>> ∀ H ∈ Turing_Machine_Deciders   
   >>>>>>>>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |   
   >>>>>>>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)   
   >>>>>>>>>>>>>>   
   >>>>>>>>>>>>>> Not exactly. A pair of otherwise identical machines that   
   >>>>>>>>>>>>>> (that are contained within the above specified set)   
   >>>>>>>>>>>>>> only differ by return value will both be wrong on the   
   >>>>>>>>>>>>>> same pathological input.   
   >>>>>>>>>>>>>   
   >>>>>>>>>>>>> You mean a pair of DIFFERENT machines. Any difference is   
   >>>>>>>>>>>>> different.   
   >>>>>>>>>>>>   
   >>>>>>>>>>>> Every decider/input pair (referenced in the above set) has a   
   >>>>>>>>>>>> corresponding decider/input pair that only differs by the   
   >>>>>>>>>>>> return   
   >>>>>>>>>>>> value of its decider.   
   >>>>>>>>>>>   
   >>>>>>>>>>> Nope.   
   >>>>>>>>>>>   
   >>>>>>>>>> ∀ H ∈ Turing_Machines_Returning_Boolean   
   >>>>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |   
   >>>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)   
   >>>>>>>>>>   
   >>>>>>>>>> Every H/TMD pair (referenced in the above set) has a   
   >>>>>>>>>> corresponding H/TMD pair that only differs by the return   
   >>>>>>>>>> value of its Boolean_TM.   
   >>>>>>>>>   
   >>>>>>>>> That isn't in the set above.   
   >>>>>>>>>   
   >>>>>>>>>>   
   >>>>>>>>>> That both of these H/TMD pairs get the wrong answer proves that   
   >>>>>>>>>> their question was incorrect because the opposite answer to the   
   >>>>>>>>>> same question is also proven to be incorrect.   
   >>>>>>>>>>   
   >>>>>>>>>>   
   >>>>>>>>> Nope, since both aren't in the set selected.   
   >>>>>>>>>   
   >>>>>>>>   
   >>>>>>>> When they are deciders that must get the correct answer both   
   >>>>>>>> of them are not in the set.   
   >>>>>>>   
   >>>>>>> *IF* they are correct decider.   
   >>>>>>>   
   >>>>>>> WHen we select from all Turing Machine Deciders, there is no   
   >>>>>>> requirement that any of them get any particular answer right.   
   >>>>>>>   
   >>>>>>> So, ALL deciders are in the set that we cycle through and apply the   
   >>>>>>> following logic to ALL of them.   
   >>>>>>>   
   >>>>>>> Each is them paired with an input that it will get wrong, and the   
   >>>>>>> existance of the input was what as just proven, the ^ template   
   >>>>>>>   
   >>>>>>>>   
   >>>>>>>> When they are Turing_Machines_Returning_Boolean the this   
   >>>>>>>> set inherently includes identical pairs that only differ   
   >>>>>>>> by return value.   
   >>>>>>>   
   >>>>>>> But in the step of select and input that they will get wrong, they   
   >>>>>>> will be givne DIFFERENT inputs.   
   >>>>>>>   
   >>>>>>>>   
   >>>>>>>>> You just don't understand what that statement is saying.   
   >>>>>>>>>   
   >>>>>>>>> I've expalined it, but it seems over you head.   
   >>>>>>>>>   
   >>>>>>>> No the problem is that you are not paying attention.   
   >>>>>>>   
   >>>>>>> No, you keep on making STUPID mistakes, like thinking that select a   
   >>>>>>> input that the machine will get wrong needs to be the same for two   
   >>>>>>> differnt machines.   
   >>>>>>>   
   >>>>>>>   
   >>>>>>>   
   >>>>>>>>   
   >>>>>>>>> For Every H, we show we can find at least one input (chosen   
   >>>>>>>>> just for   
   >>>>>>>>> that machine) that it will get wrong.   
   >>>>>>>>>   
   >>>>>>>> When we use machine templates then we can see instances of   
   >>>>>>>> the same machine that only differs by return value where both   
   >>>>>>>> get the wrong answer on the same input. By same input I mean   
   >>>>>>>> the same finite string of numerical values.   
   >>>>>>>>   
   >>>>>>>   
   >>>>>>> But if they returned differnt values, they will have different   
   >>>>>>> descriptions.   
   >>>>>>>   
   >>>>>>> Otherwise, how could a UTM get the right answer, since it only gets   
   >>>>>>> the description.   
   >>>>>>   
   >>>>>> We can get around all of this stuff by simply using this criteria:   
   >>>>>> Date 10/13/2022 11:29:23 AM   
   >>>>>> *MIT Professor Michael Sipser agreed this verbatim paragraph is   
   >>>>>> correct*   
   >>>>>> (He has neither reviewed nor agreed to anything else in this paper)   
   >>>>>> (a) If simulating halt decider H correctly simulates its input D   
   >>>>>> until H   
   >>>>>> correctly determines that its simulated D would never stop running   
   >>>>>> unless aborted then   
   >>>>>> (b) H can abort its simulation of D and correctly report that D   
   >>>>>> specifies a non-halting sequence of configurations.   
   >>>>>>   
   >>>>>> *When we apply this criteria* (elaborated above)   
   >>>>>> Will you halt if you never abort your simulation?   
   >>>>>> *Then the halting problem is conquered*   
   >>>>>>   
   >>>>>> When two different machines implementing this criteria   
   >>>>>> get different results from identical inputs then we   
      
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