Liste des Groupes | Revenir à s logic |
On 5/20/2024 2:55 AM, Mikko wrote:No, it doesn't. It is a syntax error to have the same symbol onOn 2024-05-19 14:15:51 +0000, olcott said:True(English, "a cat is an animal) is true
On 5/19/2024 9:03 AM, Mikko wrote:On 2024-05-19 13:41:56 +0000, olcott said:
On 5/19/2024 6:55 AM, Richard Damon wrote:When x is defined as True(L,x) then x is what True(L,x) is,On 5/18/24 11:47 PM, olcott wrote:True(L,x) is always a truth bearer.On 5/18/2024 6:04 PM, Richard Damon wrote:Nope, Because "This sentece is not true" can be a non-truth-bearer, but by its definition, True(L, x) can not.On 5/18/24 6:47 PM, olcott wrote:Yes we must build from mutual agreement, good.On 5/18/2024 5:22 PM, Richard Damon wrote:So, for a statement x to be false, it says that there must be a sequence of truth perserving operations that derive ~x from, right?On 5/18/24 4:00 PM, olcott wrote:My True(L,x) predicate is defined to return true or false for everyOn 5/18/2024 2:57 PM, Richard Damon wrote:Not allowed.On 5/18/24 3:46 PM, olcott wrote:The system is designed so you can ask this, yet non-truth-bearersOn 5/18/2024 12:38 PM, Richard Damon wrote:No, we can ask True(L, x) for any expression x and get an answer.On 5/18/24 1:26 PM, olcott wrote:YOU ALREADY KNOW THAT IT DOESN'TOn 5/18/2024 11:56 AM, Richard Damon wrote:And the Truth Predicate isn't allowed to "filter" out expressions.On 5/18/24 12:48 PM, olcott wrote:On 5/13/2024 9:31 PM, Richard Damon wrote:On 5/18/2024 9:32 AM, Richard Damon wrote:And thus, When True(L, p) established a sequence of truth preserving operations eminationg from ~True(L, p) by returning false, it contradicts itself. The problem is that True, in making an answer of false, has asserted that such a sequence exists.On 5/18/24 10:15 AM, olcott wrote:My True(L,x) predicate is defined to return true or false for everyOn 5/18/2024 7:43 AM, Richard Damon wrote:No, I have, but you don't understand the proof, it seems because you don't know what a "Truth Predicate" has been defined to be.No, your system contradicts itself.You have never shown this.
The most you have shown is a lack of understanding of the
Truth Teller Paradox.
finite string x on the basis of the existence of a sequence of truth
preserving operations that derive x from
> On 5/13/24 10:03 PM, olcott wrote:
>> On 5/13/2024 7:29 PM, Richard Damon wrote:
>>>
>>> Remember, p defined as ~True(L, p) ...
>>
>> Can a sequence of true preserving operations applied
>> to expressions that are stipulated to be true derive p?
> No, so True(L, p) is false
>>
>> Can a sequence of true preserving operations applied
>> to expressions that are stipulated to be true derive ~p?
>
> No, so False(L, p) is false,
>
*To help you concentrate I repeated this*
The Liar Paradox and your formalized Liar Paradox both
contradict themselves that is why they must be screened
out as type mismatch error non-truth-bearers *BEFORE THAT OCCURS*
WE HAVE BEEN OVER THIS AGAIN AND AGAIN
THE FORMAL SYSTEM USES THE TRUE AND FALSE PREDICATE
TO FILTER OUT TYPE MISMATCH ERROR
The first thing that the formal system does with any
arbitrary finite string input is see if it is a Truth-bearer:
Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
are rejected before True(L, x) is allowed to be called.
finite string x on the basis of the existence of a sequence of truth
preserving operations that derive x from
A set of finite string semantic meanings that form an accurate
verbal model of the general knowledge of the actual world that
form a finite set of finite strings that are stipulated to have
the semantic value of Boolean true.
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
So do you still say that for p defined in L as ~True(L, p) that your definition will say that True(L, p) will return false?It is the perfectly isomorphic to this:
True(English, "This sentence is not true")
when x is defined as True(L,x) then x is not a truth bearer.
in this case a truth bearer.This is known as the Truth Teller ParadoxDoesn't matter. But ir you say that "x is not a truth bearer" then,
by a truth preserving transformation, you imply that True(L,x) is
LP := ~True(L, LP) expands to ~True(~True(~True(~True(...))))
TT := True(L, TT) expands to True(True(True(True(...))))No, it doesn't, for the same reason.
not a truth bearer. As you already said that "True(L,x)" is always
a truth bearer, you imply, by another truth preeserving transformation,
that something both is and is not a truth bearer.
*Not at all*Irrelevant.
*Prolog sees the same infinite recursion and rejects it*
Les messages affichés proviennent d'usenet.