On 7/3/2026 1:53 PM, dbush wrote:False, as proven by the below which you dishonestly erased.On 7/3/2026 2:45 PM, olcott wrote:Because it ignores the input it is not any haltOn 7/3/2026 1:27 PM, dbush wrote:>On 7/3/2026 2:10 PM, olcott wrote:int Not_A_Mapping(int X)On 7/3/2026 12:10 PM, dbush wrote:>On 7/3/2026 12:52 PM, olcott wrote:Ignoring the input IS NOT A MAPPINGOn 7/3/2026 10:50 AM, dbush wrote:>On 7/3/2026 11:36 AM, olcott wrote:>On 7/3/2026 4:22 AM, Mikko wrote:>On 02/07/2026 17:51, olcott wrote:>>>
Do you know enough about C to understand that
dbush example was foolish nonsense when proposed
to show the halting problem counter-example?
It is a valid example of a C program. It was present as a part of a
claim about you, and your response was the false claim that "That
is just nonsense". Later in the discussion you offer more evidence
to support his claim.
>
His halt decider did not look at its input.
Nor is it required to. All it needs to do is map inputs to outputs.
>
So a piece of metal sitting on the ground is an automobile.
Does algorithm H map machine description X and machine input Y to an output of either 0 or 1?
>
>
A mapping is nothing more than an association of inputs to outputs,
{
return 0;
}
The above algorithm does in fact compute this mathematical mapping:
>
input | output
------------------
(any int) | 0
>>>
By construing that as a mapping is one of the screw-ups
that prevents true expressed in language from being computable.
>
My HHH applies the operational semantics of
C to its finite string input DD to correctly
determine that the DD input to HHH has no PTS
well-founded justification tree within these
operational semantics.
>
With your screwed up notion of a halt decider this
would be correct.
>
int HHH(ptr DD)
{
if (5 > 3)
return 0;
}
function at all.
The above partial halt decider meets the below requirements for all algorithms that do not halt:
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
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