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Path: ...!weretis.net!feeder9.news.weretis.net!i2pn.org!i2pn2.org!.POSTED!not-for-mail From: Richard Damon <richard@damon-family.org> Newsgroups: comp.theory Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V3 ---IGNORING ALL OTHER REPLIES Date: Sun, 16 Jun 2024 13:30:47 -0400 Organization: i2pn2 (i2pn.org) Message-ID: <v4n7g7$61l9$7@i2pn2.org> References: <v4kf3h$3h3iu$7@dont-email.me> <v4kfoa$2218$19@i2pn2.org> <v4l2mr$3l6pa$1@dont-email.me> <v4l6gg$3n5d$1@i2pn2.org> <v4l87j$3m8b0$2@dont-email.me> <v4l8jn$3n5d$3@i2pn2.org> <v4la7d$3m8b0$4@dont-email.me> <v4lan7$3n5c$2@i2pn2.org> <v4lcoo$3n4dj$3@dont-email.me> <v4leiq$3n5d$8@i2pn2.org> <v4lfu5$3rfk3$2@dont-email.me> <v4m63j$3v7mm$1@dont-email.me> <v4mmvp$1qt6$4@dont-email.me> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sun, 16 Jun 2024 17:30:47 -0000 (UTC) Injection-Info: i2pn2.org; logging-data="198313"; mail-complaints-to="usenet@i2pn2.org"; posting-account="diqKR1lalukngNWEqoq9/uFtbkm5U+w3w6FQ0yesrXg"; User-Agent: Mozilla Thunderbird X-Spam-Checker-Version: SpamAssassin 4.0.0 Content-Language: en-US In-Reply-To: <v4mmvp$1qt6$4@dont-email.me> Bytes: 6503 Lines: 124 On 6/16/24 8:48 AM, olcott wrote: > On 6/16/2024 3:00 AM, Mikko wrote: >> On 2024-06-16 01:42:29 +0000, olcott said: >> >>> On 6/15/2024 8:19 PM, Richard Damon wrote: >>>> On 6/15/24 8:48 PM, olcott wrote: >>>>> On 6/15/2024 7:13 PM, Richard Damon wrote: >>>>>> On 6/15/24 8:05 PM, olcott wrote: >>>>>>> On 6/15/2024 6:37 PM, Richard Damon wrote: >>>>>>>> On 6/15/24 7:30 PM, olcott wrote: >>>>>>>>> On 6/15/2024 6:01 PM, Richard Damon wrote: >>>>>>>>>> On 6/15/24 5:56 PM, olcott wrote: >>>>>>>>>>> On 6/15/2024 11:33 AM, Richard Damon wrote: >>>>>>>>>>>> On 6/15/24 12:22 PM, olcott wrote: >>>>>>>>>>>>> On 6/13/2024 8:24 PM, Richard Damon wrote: >>>>>>>>>>>>> > On 6/13/24 11:32 AM, olcott wrote: >>>>>>>>>>>>> >> >>>>>>>>>>>>> >> It is contingent upon you to show the exact steps of >>>>>>>>>>>>> how H computes >>>>>>>>>>>>> >> the mapping from the x86 machine language finite string >>>>>>>>>>>>> input to >>>>>>>>>>>>> >> H(D,D) using the finite string transformation rules >>>>>>>>>>>>> specified by >>>>>>>>>>>>> >> the semantics of the x86 programming language that >>>>>>>>>>>>> reaches the >>>>>>>>>>>>> >> behavior of the directly executed D(D) >>>>>>>>>>>>> >> >>>>>>>>>>>>> > >>>>>>>>>>>>> > Why? I don't claim it can. >>>>>>>>>>>>> >>>>>>>>>>>>> The first six steps of this mapping are when instructions >>>>>>>>>>>>> at the machine address range of [00000cfc] to [00000d06] >>>>>>>>>>>>> are simulated/executed. >>>>>>>>>>>>> >>>>>>>>>>>>> After that the behavior of D correctly simulated by H diverges >>>>>>>>>>>>> from the behavior of D(D) because the call to H(D,D) by D >>>>>>>>>>>>> correctly simulated by H cannot possibly return to D. >>>>>>>>>>>> >>>>>>>>>>>> Nope, the steps of D correctly simulated by H will EXACTLY >>>>>>>>>>>> match the steps of D directly executed, until H just gives >>>>>>>>>>>> up and guesses. >>>>>>>>>>>> >>>>>>>>>>> >>>>>>>>>>> When we can see that D correctly simulated by H cannot possibly >>>>>>>>>>> reach its simulated final state at machine address [00000d1d] >>>>>>>>>>> after one recursive simulation and the same applies for 2,3,...N >>>>>>>>>>> recursive simulations then we can abort the simulated input and >>>>>>>>>>> correctly report that D correctly simulated by H DOES NOT HALT. >>>>>>>>>> >>>>>>>>>> Nope. Because an aborted simulation doesn't say anything about >>>>>>>>>> Halting, >>>>>>>>>> >>>>>>>>> >>>>>>>>> It is the mathematical induction that says this. >>>>>>>>> >>>>>>>> WHAT "Mathematical Induction"? >>>>>>>> >>>>>>> >>>>>>> A proof by induction consists of two cases. The first, the base >>>>>>> case, proves the statement for n = 0 without assuming any knowledge >>>>>>> of other cases. The second case, the induction step, proves that >>>>>>> if the statement holds for any given case n = k then it must also >>>>>>> hold for the next case n = k + 1 These two steps establish that the >>>>>>> statement holds for every natural number n. >>>>>>> https://en.wikipedia.org/wiki/Mathematical_induction >>>>>> >>>>>> Ok, so you can parrot to words. >>>>>> >>>>>>> >>>>>>> It is true that after one recursive simulation of D correctly >>>>>>> simulated by H that D does not reach its simulated final state >>>>>>> at machine address [00000d1d]. >>>>>> >>>>>> Which means you consider that D has been bound to that first H, so >>>>>> you have instruciton to simulate in the call H. >>>>>> >>>>>>> >>>>>>> *We directly see this is true for every N thus no assumption needed* >>>>>>> It is true that after N recursive simulations of D correctly >>>>>>> simulated by H that D does not reach its simulated final state >>>>>>> at machine address [00000d1d]. >>>>>> >>>>>> Nope, because to do the first step, you had to bind the definition >>>>>> of the first H to D, and thus can not change it. >>>>> >>>>> So infinite sets are permanently beyond your grasp. >>>>> The above D simulated by any H has the same property >>>>> of never reaching its own simulated machine address >>>>> at [00000d1d]. >>>>> >>>>> What I mistook for dishonestly is simply a lack >>>>> of comprehension. >>>>> >>>> >>>> >>>> But it isn't an infinite set. >>>> >>> >>> Sure it is you are just clueless. >>> I mistook your ignorance for deception. >>> >>>> We don't ask an infinite set a question, or give a decider an >>>> infinite set of inputs. >>>> >>> >>> Yes we do and this is simply over your head. >>> >>> When Ĥ is applied to ⟨Ĥ⟩ >>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ >>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>> >>> The second ⊢* wildcard specifies this infinite set. >> >> As you should already know, ⊢* as used by Linz is not a wildcard. >> It is a repeated application of ⊢ without showing intermediate steps. >> > > It *is* a wild card such that the Linz template simultaneously > specifies an infinite set of machines. > of POTENTIAL machines, of which *ONE* is chosen (arbitrarily) to test. After which, it is found that the potentially infinite set of machines is actually the empty set, when you apply the requirements.