Path: ...!weretis.net!feeder9.news.weretis.net!i2pn.org!i2pn2.org!.POSTED!not-for-mail From: Richard Damon Newsgroups: comp.theory,sci.logic,comp.ai.philosophy Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V3 ---IGNORING ALL OTHER REPLIES Date: Sat, 15 Jun 2024 20:13:27 -0400 Organization: i2pn2 (i2pn.org) Message-ID: References: MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sun, 16 Jun 2024 00:13:27 -0000 (UTC) Injection-Info: i2pn2.org; logging-data="122028"; mail-complaints-to="usenet@i2pn2.org"; posting-account="diqKR1lalukngNWEqoq9/uFtbkm5U+w3w6FQ0yesrXg"; User-Agent: Mozilla Thunderbird In-Reply-To: X-Spam-Checker-Version: SpamAssassin 4.0.0 Content-Language: en-US Bytes: 5403 Lines: 103 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. If we simulate THAT input for one more step, it will see the FIRST H decide to abort and return. If you don't bind the first H, yo can't do the first step. > > _D() > [00000cfc](01) 55          push ebp > [00000cfd](02) 8bec        mov ebp,esp > [00000cff](03) 8b4508      mov eax,[ebp+08] > [00000d02](01) 50          push eax       ; push D > [00000d03](03) 8b4d08      mov ecx,[ebp+08] > [00000d06](01) 51          push ecx       ; push D > [00000d07](05) e800feffff  call 00000b0c  ; call H > [00000d0c](03) 83c408      add esp,+08 > [00000d0f](02) 85c0        test eax,eax > [00000d11](02) 7404        jz 00000d17 > [00000d13](02) 33c0        xor eax,eax > [00000d15](02) eb05        jmp 00000d1c > [00000d17](05) b801000000  mov eax,00000001 > [00000d1c](01) 5d          pop ebp > [00000d1d](01) c3          ret > Size in bytes:(0034) [00000d1d] > > >> You haven't shown the required pieces for an inductive proof. >> >> I doubt you even know what you need to do, let alone be able to do it. > >