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On 5/20/24 2:59 PM, olcott wrote:That is the same thing as saying thatOn 5/19/2024 6:30 PM, Richard Damon wrote:Irrelvent.On 5/19/24 4:12 PM, olcott wrote:>On 5/19/2024 12:17 PM, Richard Damon wrote:>On 5/19/24 9:41 AM, olcott wrote:>>>
True(L,x) is always a truth bearer.
when x is defined as True(L,x) then x is not a truth bearer.
So, x being DEFINED to be a certain sentence doesn't make x to have the same meaning as the sentence itself?
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What does it mean to define a name to a given sentence, if not that such a name referes to exactly that sentence?
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p = ~True(L,p) // p is not a truth bearer because its refers to itself
Then ~True(L,p) can't be a truth beared as they are the SAME STATEMENT, just using different "names".
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Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
p = ~True(L,p) Truthbearer(L,p) is false
q = ~True(L,p) Truthbearer(L,q) is true
If Truthbearer(L, p) is FALSE, and since p is just a NAME for the statement ~True(L, p), that means that True(L. p) is not a truth bearer and True has failed to be the required truth predicate.
If you are defining your "=" symbol to be "is defined as" so the left side is now a name for the right side, you statement above just PROVES that your logic system is inconsistant as the same expression (with just different names) has contradicory values.ϕ(x) there is a sentence ψ such that S ⊢ ψ ↔ ϕ⟨ψ⟩.
You are just showing you utter lack of understanding of the fundamentals of Formal Logic.
Likewise with ~True(L, ~True(L, p)) where p is defined as ~True(L, p)Right, that is a sentence about another sentence (that is part of itself)>>
Just like (with context) YOU can be refered to a PO, Peter, Peter Olcott or Olcott, and all the reference get to the exact same entity, so any "name" for the express
>True(L,p) is false>
True(L,~p) is false
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So since True(L, p) is false, then ~True(L, p) is true.
>~True(True(L,p)) is true and is referring to the p that refers>
to itself it is not referring to its own self.
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*ONE LEVEL OF INDIRECT REFERENCE MAKES ALL THE DIFFERENCE*
Why add the indirection? p is the NAME of the statement, which means exactly the same thing as the statement itself.
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p = ~True(L,p)
does not mean that same thing as True(L, ~True(L,p))
The above ~True(L, p) has another ~True(L,p) embedded in p.
>Is the definition of an English word one level LESS of indirection than the word itself?>
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This sentence is not true("This sentence is not true") is true.
p defined as ~True(L, p) isn't a sentence refering to ~True(L, p), it is assigning a name to the sentence to allow OTHER sentences to refer to it by name,Yet when p refers to its own name this creates infinite recursion.
When LP refers to its own name this creates infinite recursion.>Nope.I don't think you understand what it means to define something.>
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x := y means x is defined to be another name for y
https://en.wikipedia.org/wiki/List_of_logic_symbols
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LP := ~True(L, LP)
specifies ~True(~True(~True(~True(~True(...)))))
It means that LP is defined to be the sentence ~True(L, LP)True(English, "this sentence is not true") is false
replacing the LP in the sentence with a copy of LP IS a level of indirection, so you can get the infinite expansion if you keep or derefencing the reference in the statement.
>Which isn't a valid proof in a formal system. You seem to think Formal System are a loosy goosy with proofs as Philosophy."Definition by example" is worse than "Proof by example", at least proof by example can be correct if the assertion is that there exists, and not for all.>
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A simpler isomorphism of the same thing is proof by analogy.
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LP := ~True(L, LP) means that every instance of LPNope, it is equivelent to that, but doesn't SPECIFY that.A level of indirection:>
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p: "This sentence is true", which is exactly the same as "p is true" since "this sentence" IS p
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p := True(L,p)
specifies True(True(True(True(True(...)))))
As I said above that is expanding levels of indirecction.Nothing can handle "some kind of infinite structure."
>Right, because prolog can't handle any levels of self referencing, and thus is not suitable for logic that can do that.
*Prolog sees the same infinite recursion and rejects it*
?- TT = true(TT).
TT = true(TT).
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?- unify_with_occurs_check(TT, true(TT)).
false.
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You have been told this, but don't seem to understand it. My guess is you can't understand any logic more complicated than what Prolog handles, so don't realize how much it just doesn't handle.No the whole problem seems to be that you simply don't
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