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 8:10 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. Correction: it correctly reports the halt status of any algorithm that *does not* 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 >