Re: The philosophy of logic reformulates existing ideas on a new basis --- infallibly correct

On 11/13/24 11:44 AM, olcott wrote:

On 11/13/2024 5:52 AM, Richard Damon wrote:

On 11/12/24 11:37 PM, olcott wrote:

On 11/11/2024 9:06 AM, Richard Damon wrote:

On 11/10/24 5:01 PM, olcott wrote:

On 11/10/2024 2:39 PM, joes wrote:

Am Sun, 10 Nov 2024 14:07:44 -0600 schrieb olcott:

On 11/10/2024 1:13 PM, Richard Damon wrote:

On 11/10/24 10:11 AM, olcott wrote:

On 11/10/2024 4:03 AM, Alan Mackenzie wrote:

In comp.theory olcott <polcott333@gmail.com> wrote:

On 11/9/2024 4:28 PM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 11/9/2024 3:45 PM, Alan Mackenzie wrote:

>

Sorry, but until you actually and formally fully define your logic
system, you can't start using it.

When C is a necessary consequence of the Haskell Curry elementary
theorems of L (Thus stipulated to be true in L) then and only then is C
is True in L.
This simple change does get rid of incompleteness because Incomplete(L)
is superseded and replaced by Incorrect(L,x).

I still can’t see how this makes ~C provable.
>

>
If C is not provable it is merely rejected as incorrect
not used as any basis to determine that L is incomplete.
>
For many reasons: "A sequence of truth preserving operations"
is a much better term than the term "provable".
>

>
But since there exist statements that are True but not Provable. except by your incorrect definition of Provable, your logic is just broken.
>

>
There cannot possibly be any expressions of language that
are true in L that are not determined to be true on the
basis of applying a sequence of truth preserving operations
in L to Haskell_Curry_Elementary_Theorems in L.
>

>
Right, but there can be expressions of language that are true in L by an INFINITE sequence of truth-preserving operations that are not provable which needs a FINITE sequence of truth-preserving operations.
>

 If it is impossible to show that x is true in L and impossible
to show that ~x is true in L then x in not a truth bearer in L
and L is by no means in any way incomplete.
 x = "This sentence is not true"
True(English, x) == false. True(English, ~x) == false.
 

But if x can only be shown to be true by an INFINTE chain of steps, which thus do not form the FINITE chain needed for a proof, L meets the requirements for incompleteness.
And you meed the requirements for a blantant liar, as you KNOW that proofs in standard logic are finite, while truth is allowed to be infinite.
So, you are just hoisted on your own strawman and burning your soul in effigy.