Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

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Skepdick
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Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by Skepdick »

PeteOlcott wrote: Fri May 13, 2022 6:10 am So you find an unconventional unpublished proof, that may or may not be correct, (no peer review).
None-The-Less I have defeated ALL the conventional proofs.
This post on contains a link to a YouTube video. You really really should watch that video as an introduction to constructivism.

https://cs.stackexchange.com/questions/ ... lem-in-coq

In a nutshell you are doing proof by negation.

You produce H which negates P.
I produce P1 which negates H.
You produce H1 which negates P1.
I produce P2 which negates H1
...
Negation all the way down.

That's how falsification works in science.
PeteOlcott
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Joined: Mon Jul 25, 2016 6:55 pm

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by PeteOlcott »

Skepdick wrote: Fri May 13, 2022 6:29 am
PeteOlcott wrote: Fri May 13, 2022 6:10 am So you find an unconventional unpublished proof, that may or may not be correct, (no peer review).
None-The-Less I have defeated ALL the conventional proofs.
This post on contains a link to a YouTube video. You really really should watch that video as an introduction to constructivism.

https://cs.stackexchange.com/questions/ ... lem-in-coq

In a nutshell you are doing proof by negation.

You produce H which negates P.
I produce P1 which negates H.
You produce H1 which negates P1.
I produce P2 which negates H1
...
Negation all the way down.

That's how falsification works in science.
You produce H which negates P.
I produce P1 which negates H.
I abort P1 and report non-halting before P1 gets to its negation.
(my current proof)
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by Skepdick »

PeteOlcott wrote: Sat May 14, 2022 3:57 pm I abort P1 and report non-halting before P1 gets to its negation.
(my current proof)
Oh, OK!

So I can do the exact same thing then?

I simulate H1 in P2.
I abort H1 before it gets to its negation of P1 halting.

If H1 proves that P1 doesn't halt, then P2 proves that H1 doesn't prove that P1 fails to halt.

Game theory 101 - strategy-stealing strategy.
PeteOlcott
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Joined: Mon Jul 25, 2016 6:55 pm

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by PeteOlcott »

Skepdick wrote: Mon May 16, 2022 9:07 am
PeteOlcott wrote: Sat May 14, 2022 3:57 pm I abort P1 and report non-halting before P1 gets to its negation.
(my current proof)
Oh, OK!

So I can do the exact same thing then?

I simulate H1 in P2.
I abort H1 before it gets to its negation of P1 halting.

If H1 proves that P1 doesn't halt, then P2 proves that H1 doesn't prove that P1 fails to halt.

Game theory 101 - strategy-stealing strategy.
H(P,P)==0 and H1(P,P)==1 are provably correct
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by Skepdick »

PeteOlcott wrote: Mon May 16, 2022 7:14 pm H(P,P)==0 and H1(P,P)==1 are provably correct
P1(H(P,P)) == not(H(P,P)==0) is also provably correct.
P2(H1(P1, P1)) == not(H1(P1, P1) == 1) is also provably correct.
PeteOlcott
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Joined: Mon Jul 25, 2016 6:55 pm

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by PeteOlcott »

Skepdick wrote: Mon May 16, 2022 7:20 pm
PeteOlcott wrote: Mon May 16, 2022 7:14 pm H(P,P)==0 and H1(P,P)==1 are provably correct
P1(H(P,P)) == not(H(P,P)==0) is also provably correct.
P2(H1(P1, P1)) == not(H1(P1, P1) == 1) is also provably correct.
There is no P1. There is only this literal string of x86 machine language:

558bec8b4508508b4d0851e840feffff83c40885c07402ebfe5dc3

Code: Select all

_P()
[000009d6](01) 55         push ebp
[000009d7](02) 8bec       mov ebp,esp
[000009d9](03) 8b4508     mov eax,[ebp+08]
[000009dc](01) 50         push eax         // push P
[000009dd](03) 8b4d08     mov ecx,[ebp+08]
[000009e0](01) 51         push ecx         // push P
[000009e1](05) e840feffff call 00000826    // call H
[000009e6](03) 83c408     add esp,+08
[000009e9](02) 85c0       test eax,eax
[000009eb](02) 7402       jz 000009ef
[000009ed](02) ebfe       jmp 000009ed
[000009ef](01) 5d         pop ebp
[000009f0](01) c3         ret              // Final state
Size in bytes:(0027) [000009f0]
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by Skepdick »

PeteOlcott wrote: Mon May 16, 2022 7:14 pm H(P,P)==0 and H1(P,P)==1 are provably correct
But seriously. There's nothing to it. H(P,P)==0 is just a proposition valid. It doesn't mean anything until evaluated.

If H(P,P) returns 1 then the proposition is false.
If H(P,P) returns 0 then the proposition is true.
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by Skepdick »

PeteOlcott wrote: Mon May 16, 2022 7:31 pm There is no P1.

There is no H either.
PeteOlcott
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Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by PeteOlcott »

Skepdick wrote: Mon May 16, 2022 7:31 pm
PeteOlcott wrote: Mon May 16, 2022 7:14 pm H(P,P)==0 and H1(P,P)==1 are provably correct
But seriously. There's nothing to it. H(P,P)==0 is just a proposition valid. It doesn't mean anything until evaluated.

If H(P,P) returns 1 then the proposition is false.
If H(P,P) returns 0 then the proposition is true.
That is a stupid thing to say everyone knows that 0 is false and non-zero is true in C.

H(P,P)==0 (meaning input is non-halting) and
H1(P,P)==1 (meaning input halts)

H(P,P) that correctly decides its input is fully operational code.
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by Skepdick »

PeteOlcott wrote: Mon May 16, 2022 7:53 pm That is a stupid thing to say everyone knows that 0 is false and non-zero is true in C.

H(P,P)==0 (meaning input is non-halting) and
H1(P,P)==1 (meaning input halts)

H(P,P) that correctly decides its input is fully operational code.
Are you stupid? == is a predicate! An operation with arity 2.

==(H(P,P), 0) returns True if H(P,P) is equal to 0.
==(H(P,P), 0) returns False if H(P,P) is equal to 0.
==(H(P,P), 1) returns True if H(P,P) is equal to 1.
==(H(P,P), 1) returns True if H(P,P) is equal to 0.

Code: Select all

#include <stdio.h>
#include <assert.h>

int main()
{
    int x = 1;
    assert(x==1); //This assertion passes
    assert(x==0); //This assertion aborts the program
    return 0;
}
PeteOlcott
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Joined: Mon Jul 25, 2016 6:55 pm

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by PeteOlcott »

Skepdick wrote: Mon May 16, 2022 7:58 pm
PeteOlcott wrote: Mon May 16, 2022 7:53 pm That is a stupid thing to say everyone knows that 0 is false and non-zero is true in C.

H(P,P)==0 (meaning input is non-halting) and
H1(P,P)==1 (meaning input halts)

H(P,P) that correctly decides its input is fully operational code.
Are you stupid? == is a predicate. An assertion!

==(H(P,P), 0) returns True if H(P,P) is equal to 0.
==(H(P,P), 0) returns False if H(P,P) is equal to 0.
==(H(P,P), 1) returns True if H(P,P) is equal to 1.
==(H(P,P), 1) returns True if H(P,P) is equal to 0.

Code: Select all

#include <stdio.h>
#include <assert.h>

int main()
{
    int x = 1;
    assert(x==1); //This assertion passes
    assert(x==0); //This assertion aborts the program
    return 0;
}
bool H(P,P) is defined to return 0 meaning that its input is non-halting.
If you disagree with this you are a a liar.
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by Skepdick »

PeteOlcott wrote: Mon May 16, 2022 8:07 pm bool H(P,P) is defined to return 0 meaning that its input is non-halting.
H is defined as a Boolean! Meaning that H can return a 1 or a 0 !

Which, in turn means that H(P,P) == 0 could be true when H returns 0; or false when H returns 1 !
PeteOlcott wrote: Mon May 16, 2022 8:07 pm If you disagree with this you are a a liar.
OK. So you are stupid.

You don't even understand the difference between the expression "1==0" which returns a Boolean,
and an assertion 1==0 which aborts the execution of the program because the assertion is false.

Here is the code.

Code: Select all

#include <stdio.h>
#include <assert.h>

int main()
{
    0==1; 
    assert(1==0);
    return 0;
}
Here is the warning you get when you compile the code

Code: Select all

➜  ~ gcc main.c
main.c:6:6: warning: equality comparison result unused [-Wunused-comparison]
    0==1;
    ~^~~
1 warning generated.
Here is the runtime error you get when you run the code (which contains warnings)

Code: Select all

➜  ~ ./a.out
Assertion failed: (1==0), function main, file main.c, line 7.
[1]    24650 abort      ./a.out
And here is the compile-time error you get when you make the compiler strict!

Code: Select all

➜  ~ gcc -Werror main.c
main.c:6:6: error: equality comparison result unused [-Werror,-Wunused-comparison]
    0==1;
    ~^~~
1 error generated.
PeteOlcott
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Joined: Mon Jul 25, 2016 6:55 pm

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by PeteOlcott »

H(P,P)==0 means that when H is applied to this fixed string of x86 machine code:
558bec8b4508508b4d0851e840feffff83c40885c07402ebfe5dc3
H correctly determines that it never halts.

All the double talk strawman error cannot change this simple fact.
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by Skepdick »

PeteOlcott wrote: Mon May 16, 2022 8:25 pm H(P,P)==0 means that when H is applied to this fixed string of x86 machine code:
558bec8b4508508b4d0851e840feffff83c40885c07402ebfe5dc3
H correctly determines that it never halts.

All the double talk strawman error cannot change this simple fact.
Bullshit.

If H always halts (you claim it does) then H(P,P)==0 means either true; or false.

Why are you having such a hard time understanding that 0 == 1 is a boolean, not a constant?

Code: Select all

>>> type(0 == 1)
<class 'bool'>
PeteOlcott
Posts: 1062
Joined: Mon Jul 25, 2016 6:55 pm

Re: Proof that H(P,P)==0 is correct [ refuting the halting problem proofs ]

Post by PeteOlcott »

Skepdick wrote: Mon May 16, 2022 8:31 pm
PeteOlcott wrote: Mon May 16, 2022 8:25 pm H(P,P)==0 means that when H is applied to this fixed string of x86 machine code:
558bec8b4508508b4d0851e840feffff83c40885c07402ebfe5dc3
H correctly determines that it never halts.

All the double talk strawman error cannot change this simple fact.
Bullshit.

If H always halts (you claim it does) then H(P,P)==0 means either true; or false.
H(P,P) is answering the question: Does the input to H(P,P) halt?
The answer is 0,No, false, does not halt.
Why is true and false so hard for you?
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