On 5/20/2024 10:37 PM, Richard Damon wrote:
On 5/20/24 10:56 PM, olcott wrote:
On 5/20/2024 9:24 PM, Richard Damon wrote:
On 5/20/24 9:54 PM, olcott wrote:
On 5/20/2024 7:57 PM, Richard Damon wrote:
On 5/20/24 2:59 PM, olcott wrote:
On 5/19/2024 6:30 PM, Richard Damon wrote:
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:
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True(L,x) is always a truth bearer.
when x is defined as True(L,x) then x is not a truth bearer.
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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
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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
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Irrelvent.
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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.
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That is the same thing as saying that
True(English, "this sentence is not true") is false
proves that True(L,x) is not a truthbearer.
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Nope, why do you say that?
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What logic are you even TRYING to use to get there?
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I think you don't understand what defining a label to represent a statement means.
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I did not said the above part exactly precisely to address
your objection.
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p is defined as ~True(L,p)
LP is defined as "this sentence is not true" in English.
Thus True(L,p) ≡ True(English,LP) and
Thus True(L,~p) ≡ True(English,~LP)
So, you admit that you did not answer the problem.
And that you think Strawmen and Red Herring are valid forms of logic.
How does p defined as ~True(L, p) NOT generate the shown contradiction when you begin by saying True(L, p) must not be true (and thus false) because p has not chain to truthbears?
p := ~True(L, p) is false
p := ~True(L, ~p) is false
p is tossed out on its ass as a type mismatch error for every system
of bivalent logic before it gets any chance to be evaluated in any
other way.
If your gas can for you lawnmower is filled with water
do you use it anyway or dump it out?
You are just showing that you think it is ok for logical system to have contradictions in them.
You are failing to pay enough attention or forgetting
what I told you even after telling you many times.
Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
p defined as ~True(L, p)
if (~Truthbearer(L,p))
printf("%s is rejected as not a truth bearer\n", "p");
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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.
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You are just showing you utter lack of understanding of the fundamentals of Formal Logic.
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ϕ(x) there is a sentence ψ such that S ⊢ ψ ↔ ϕ⟨ψ⟩.
The sentence ψ is of course not self-referential in a strict sense, but mathematically it behaves like one. https://plato.stanford.edu/entries/self-reference/#ConSemPar
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So? Can you show that it is NOT true? or is it just that you don't want it to be true, so you assume it isn't?
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defined as is the way to go.
Which mean?
p defined as ~True(L, p)
Is much better than the incorrect conventional way: p ↔ ~True(L, p)
And what does it have to do with the original statement?
Remember, if your goal is to just show that conventonal logic is just broken, you are going to need to make a much more convincing arguement to scrap it, unless you have a FULLY DEVELOPED alternative that does better.
Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
Expressions that are {true on the basis of meaning} are ONLY
(a) A set of finite string semantic meanings that form an accurate
model of the general knowledge of the actual world.
(b) Expressions derived by applying truth preserving operations to (a).
The above algorithm specifies True(L,x) and True(L,~x).
Just remember, once you throw out the foundations, you need to start from a brand new foundation, and unless you have been lying about your prognossis, and sand-bagging about your logical abilities, your chance of actually proving somethiing like that is just about zero.
In other words you totally forgot that you already understood
the algorithm.
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No what it shows is that formal logic gets the wrong answer because
formal logic does not evaluate actual self-reference.
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No, you don't understand what you are talking about.
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Formal logic NEVER EVER gets to
epistemological antinomies ARE NOT TRUTH BEARERS
Of course it does.
You just don't understand what you are reading.
In fact, Tarski points out the BECAUSE he can show that the existance of a Truth Primative forces an epistemological antinomy to have a truth value, that there can not be an existing Truth Primative.
YOU just don't understand logic,
I understand that the received view is proven to be incorrect on the basis of its incoherence. The system of (a) and (b) is self-evidently
correct.
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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
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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.
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~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*
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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.
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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.
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Right, that is a sentence about another sentence (that is part of itself)
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Likewise with ~True(L, ~True(L, p)) where p is defined as ~True(L, p)
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So? Yes ~True(L, ~True(L, p)) IS a different sentence than ~True(L, p) even with p defined a ~True(L, p), BUT they are logically connected as the first follows as a consequence of the second and the definition of p.
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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,
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Yet when p refers to its own name this creates infinite recursion.
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So? What's wrong with that?
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Sure any programs that get stuck in infinite loops are a feature that
everyone likes even when it means that payroll is two weeks late and
you missed your mortgage payment.
Which has nothing to do with the Halting Problem.
You said there is nothing wrong with loops and I countered
with a loop that could force you to skip paying your mortgage.
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Note, it is recursion that doesn't HAVE to be followed. You seem to be stuck at counting the fingers level math, while trying to talk about trigonometry.
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Any expression "standing for some kind of infinite structure."
CANNOT BE EVALUATED THUS CANNOT POSSIBLY BE A TRUTH BEARER
THUS <IS> A TYPE MISMATCH ERROR FOR EVERY SYSTEM OF BIVALENT LOGIC
So, I guess you don't beleive in mathematics.
Those are not required to be derived from a set of truth
preserving operations that have a cycle in the directed
graph of their evaluation sequence.
And the value of Pi doesn't exist, or the square root of 2.
You are just incapable of understanding how infinities CAN work.
There is no NEED to expand the reference loop to infinity, so that isn't actually a problem.
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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(...)))))
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Nope.
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When LP refers to its own name this creates infinite recursion.
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So? As I said, it doesn't HAVE to be fully expanded, as each level is doing a logical step of indirection
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It means that LP is defined to be the sentence ~True(L, LP)
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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.
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"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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Which isn't a valid proof in a formal system. You seem to think Formal System are a loosy goosy with proofs as Philosophy.
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True(English, "this sentence is not true") is false
Is 100% perfectly isomorphic to its formalized version
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LP is defined as ~True(L, LP)
True(L, LP) is false
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Nope. Because "this sentence" refers to the statement in quotes, not the logical statement using True.
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The English is formalized as LP is defined as ~True(L, LP)
before it is analyzed.
Nope, because the English doesn't carry the meaning of being a Truth Predicate.
In other words
True(English, "Puppies are fifteen story office buildings")
is not false?
But, since you don't seem to understand what that means, you can't tell the difference, but it proves your own ignorance to make the claim.
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It is merely easier to see that "this sentence is not true"
cannot be true because that makes it false and
can't be false because that makes it true.
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And it is a different sentence.
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No it is not.
The English is formalized as
LP is defined as ~True(L, LP) before it is analyzed.
Nope, You can't make that claim.
I am correct and you can't show otherwise.
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LP is defined as ~True(L, LP)
works this same yet yet it is not as intuitive.
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You are right that it causes problems, and the problem it causes is that it shows that the True Predicate can not exist.
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Not at all.
It shows that NON truth bearers must be rejected as
a type mismatch error for any system of bivalent logic.
Which isn't allowed.
I had a typo : NON truth bearers must be rejected
Truthbearer(L,x) ≡ (True(L,x) ∨ False(L,x))
You seem to have this problem with things defined to work on ALL statements expressable in the languge.
My system recognizes and reject epistemological antinomies.
It is DEFINED how the Truth predicate is to work on non-truth bearers, and that to return the false value.
Truthbearer(L,x) ≡ (True(L,x) ∨ False(L,x))
It is basically defined similar to Sipser Decider, in that it turns "non-answers" into a defined answer, and that requirement is what make it not possible, but that requirement is a fundamental part of the problem.
Are there a sequence of truth preserving operations that derive
x from
(a) A set of finite string semantic meanings that form an accurate
model of the general knowledge of the actual world.
No means x is not true.
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So we see that the above is a correct formalization
of the English and that gives us the cognitive leverage
of intuition.
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Nope, can't because the English sentence doesn't attach a "name" to the whole expression.
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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(...)))))
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Nope, it is equivelent to that, but doesn't SPECIFY that.
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LP := ~True(L, LP) means that every instance of LP
in the RHS is the same as the RHS.
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Clocksin & Mellish say this same thing.
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BEGIN:(Clocksin & Mellish 2003:254)
Finally, a note about how Prolog matching sometimes differs from the
unification used in Resolution. Most Prolog systems will allow you to
satisfy goals like:
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And how Prolog does it is irrelevent,
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Not at all.
Prolog sees that LP is defined as ~True(LP) is nonsense
and rejects it.
And thus proves that it can't handle the logic.
*THE FREAKING INPUT IS FREAKING WRONG*
*THE FREAKING INPUT IS FREAKING WRONG*
*THE FREAKING INPUT IS FREAKING WRONG*
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
True on the basis of meaning
Re: True on the basis of meaning