Re: The simple essence of Proof Theoretic Semantics
dbush <[email protected]>
| Newsgroups | sci.logic,comp.theory,comp.ai.philosophy,sci.math |
|---|---|
| Organization | A noiseless patient Spider |
| Message-ID | <[email protected]> |
On 7/3/2026 10:02 PM, olcott wrote: > On 7/3/2026 7:10 PM, dbush wrote: >> On 7/3/2026 6:37 PM, olcott wrote: >>> On 7/3/2026 1:47 PM, André G. Isaak wrote: >>>> On 2026-07-03 12:36, olcott wrote: >>>>> On 7/3/2026 1:18 PM, dbush wrote: >>>> >>>>>> If an algorithm takes an input and produces an output, that is by >>>>>> definition a mapping. >>>>> That only proves that the definition is incoherent. >>>>> The coherent way that it actually works is that >>>>> inputs are transformed into outputs by applying >>>>> finite string transformation rules to inputs to >>>>> derive outputs. >>>> >>>> Apparently you don't understand the difference between a mapping and >>>> an algorithm. They are two different things. >>>> >>>> André >>>> >>> >>> A function that ignores its input and only returns 0 >>> is not any sort of halt function. >>> >> >> It is a partial halt decider that correctly reports the halt status of >> any algorithm that halts when executed directly and incorrectly >> reports the halt status of algorithms that halt when executed directly. >> >> If you disagree, point out exactly which part of the below >> requirements is violated in doing so. If you dishonestly trim this, >> it will be taken as your official, on-the-record admission that the >> below requirements are satisfied for the subset of algorithms that >> halt when executed directly. >> >> >> Given any algorithm (i.e. a fixed immutable sequence of instructions) >> X described as <X> with input Y: >> >> A solution to the halting problem is an algorithm H that computes the >> following mapping: >> >> (<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed >> directly >> > > A actual halt function must compute the mapping Shown above.