Refuting Gödel's 1931 Incompleteness Theorem in one sentence

[PeteOlcott]

I figured out how to solve the halting problem.

This has nothing to do with Turing completeness.

If you are not solving the halting problem in the context of Turing-complete languages then you aren't solving the halting problem.

You are solving something else.

OK that aspect is relevant. My solution works on Turing complete machines, none of which actually exist or could possibly ever exist.

For you to presume that solving the halting problem requires and oracle machine is an unjustified presumption.

[Univalence]

OK that aspect is relevant. My solution works on Turing complete machines, none of which actually exist or could possibly ever exist.

Read that sentence again. Carefully. You are solving a "problem" on a non-existing architecture.

For you to presume that solving the halting problem requires and oracle machine is an unjustified presumption.

Strawman. That which can solve the general case of the halting problem IS an Oracle machine.

[PeteOlcott]

I have refuted one instance of the general case

This is an oxymoron.

providing the means to refute every other case using conventional programming and my refutation pattern.

Total functional programming languages already do that. They detect almost all non-halting functions.

Almost.

http://liarparadox.org/Peter_Linz_HP(Pages_318-319).pdf

Peter Linz provides the state transition specification of every conventional halting problem
proof of H^ defined in terms of H. This pattern is the proposed totally unbreakable pattern
of every HP proof. I broke this pattern. H actually decides halting for one instance of (H^, H^).

Now that the impenetrable wall has a hole punched through it, the rest of the wall
can be torn down.

[Univalence]

I broke this pattern. H actually decides halting for one instance of (H^, H^).

You keep saying it without producing an actual algorithm. Just some squiggles on paper that somebody is supposed to take seriously.

I am starting to think you are just trying to bullshit your way through.

[PeteOlcott]

I broke this pattern. H actually decides halting for one instance of (H^, H^).

You keep saying it without producing an actual algorithm. Just some squiggles on paper that somebody is supposed to take seriously.

I am starting to think you are just trying to bullshit your way through.

Not at all. I am just establishing the basis by which my work should be evaluated.
If my single counter-example does break the pattern then this single counter
example does show how to solve the halting problem.

I have to finish my friend's disability case before I can actually focus on completing
the coding of the UTM interpreter.

[Univalence]

Not at all. I am just establishing the basis by which my work should be evaluated.
If my single counter-example does break the pattern then this single counter
example does show how to solve the halting problem.

If your single counter example is implemented in Turing-complete language and I can provide you with malformed input which causes your algorithm to enter an infinite loop (e.g your own "proof" becomes undecidable).

Then would you agree that your "counter example" is actually evidence for and not against the halting problem's validity?

[PeteOlcott]

Not at all. I am just establishing the basis by which my work should be evaluated.
If my single counter-example does break the pattern then this single counter
example does show how to solve the halting problem.

If your single counter example is implemented in Turing-complete language and I can provide you with malformed input which causes your algorithm to enter an infinite loop (e.g your own "proof" becomes undecidable).

Then would you agree that your "counter example" is actually evidence for and not against the halting problem's validity?

The reason that I am not just providing the TM source code for this single counter-example and
instead creating a full UTM interpreter is so that I can quickly write more TM code to
decide any other instance and then immediately prove that it works with a full execution
trace.

I expect that your counter-example is something like If Provable(Goldbach Conjecture) then halt else loop.

All counter-examples like this are simply deductively unsound because they depend on information
that does not currently exist. If they depended on information that cannot possibly exist, then that
would be another example of deductively unsound.

[Univalence]

The reason that I am not just providing the TM source code for this single counter-example and
instead creating a full UTM interpreter is so that I can quickly write more TM code to
decide any other instance and then immediately prove that it works with a full execution
trace.

1. You can't have a "full" execution trace of a program that has not yet terminated
2. The strategy you suggest to addressing the issue is akin to an arms race.

You are forever solving edge cases, and I am forever finding a way to exploit your "fix". But you have no universal solution.

I expect that your counter-example is something like If Provable(Goldbach Conjecture) then halt else loop.

My counter-example will be dependant on your "solution". Somewhere in your code you WILL have a function capable of parsing/executing code, otherwise your system is not Turing-complete.

Be it the WFF() function, your type-checker. If your system is Turing-complete I can make it behave any way I want.

My strategy is to keep preventing your own "counter example" from halting.

[PeteOlcott]

The reason that I am not just providing the TM source code for this single counter-example and
instead creating a full UTM interpreter is so that I can quickly write more TM code to
decide any other instance and then immediately prove that it works with a full execution
trace.

1. You can't have a "full" execution trace of a program that has not yet terminated

Sure everyone knows that every correct halt decider must itself always get stuck in an infinite loop.

[Univalence]

All counter-examples like this are simply deductively unsound because they depend on information
that does not currently exist. If they depended on information that cannot possibly exist, then that
would be another example of deductively unsound.

Nonsense. This is the domain of linear logic and partial functions.

The information needs not exist. Only a strategy for obtaining the information needs exist.

A recursive function for determining the Nth Fibonacci number does not "know" the answer at this very moment.
It only knows how to get the answer given N time-cycles.

[Univalence]

Sure everyone knows that every correct halt decider must itself always get stuck in an infinite loop.

Then how have you solved the Halting Problem if the decider itself may or may not halt?

[PeteOlcott]

Sure everyone knows that every correct halt decider must itself always get stuck in an infinite loop.

Then how have you solved the Halting Problem if the decider itself may or may not halt?

You can't detect sarcasm?

[Univalence]

You can't detect sarcasm?

Not from a person who claims to have invented a universal, always-halting, Turing-complete decider.

[PeteOlcott]

You can't detect sarcasm?

Not from a person who claims to have invented a universal, always-halting, Turing-complete decider.

I already said that is only decides one single instance of the "impossibly decidable" generic halting problem proof pattern.

[Univalence]

I already said that is only decides one single instance of the "impossibly decidable" generic halting problem proof pattern.

And I already told you that your algorithm can't decide its own undecidability if:
1. Your language is Turing-complete
2. You allow me to craft the input.

What you have is a work-around, not a solution.

Otherwise, I also solved the halting problem:

Code: Select all

def halt(x):
  return True

All hardware fails -> all algorithms halt.