Re: The simple essence of Proof Theoretic Semantics

On 7/3/2026 2:36 PM, olcott wrote:
> On 7/3/2026 1:18 PM, dbush wrote:
>> On 7/3/2026 2:10 PM, olcott wrote:
>>> On 7/3/2026 12:10 PM, dbush wrote:
>>>> On 7/3/2026 12:52 PM, olcott wrote:
>>>>> On 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?
>>>>
>>> Ignoring the input IS NOT A MAPPING
>>>
>>
>> If an algorithm takes an input and produces an output, that is by 
>> definition a mapping. 
> That only proves that the definition is incoherent.

There is nothing incoherent about that.  But if it makes you happy let's 
instead use the below algorithm H1 as a potential halt decider and 
algorithm D1 which is designed via the counter-example template to make 
algorithm H1 get the wrong answer.

void D(ptr *I)
{
      // algorithm D1; input: I
      ptr *X = D;
      ptr *Y = I;
      int result;
      {
          // algorithm H1; inputs: X,Y
          result = 0 + (X-X) + (Y-Y);
      }
      if (result == 1) {
          while (1);
      }
}

int H(ptr *X, ptr *Y)
{
      int result;
      {
          // algorithm H1; inputs: X,Y
          result = 0 + (X-X) + (Y-Y);
      }
      return result;
}

There, now the input is not ignored.