In ordinary mathematical logic, contradictions are syntactical, not requiring assignment of truth values. Meanwhile, as far as I can tell, your layer logic is described primarily semantically in terms of truth values; I don't know the syntax of whatever proof system you have in mind, so I can't evaluate the means by which you would prevent (syntactical) contradictions. You could assert that provability entails soundness, but we need to prove that, not just assert it, and you can't prove it without first stating what the proof system is. — TonesInDeepFreeze
I would guess that layer logic does disprove contradictions. That is, layer logic disproves all formulas of the form 'P & ~P' (where they "reside" (or\e whatever way you say it) in the same level [should be layer]). — TonesInDeepFreeze
what you say is that there are three truth values and that statements are evaluated at different levels [should be layers]. You haven't given even the starting point: description of evaluation of truth and falsehood for atomic sentences, compound sentences, and quantificational sentences. — TonesInDeepFreeze
how I handle the proof of the halting problem — Trestone
with layer logic we have to add layers if a program has to give a value/result:
A given program halts or not in layer k for given input data. — Trestone
my earlier handling of Cantor´s diagonalization and proof in layer logic — Trestone
(t marks the layers, W(x,t) ist the truth value of x in layer t, -w stands for „not true“ or „false“
ther value „undefined“ I left out to make things easier).
Be M a set and P(M) its power set and F: M -> P(M) a bijection between them (in layer d)
Then the set A is defined with W(x e A, t+1) = w := if ( W(x e M,t)=w and W(x e F(x),t)=-w ) — Trestone
F: M -> P(M) a bijection — Trestone
the proof about the power set can be similary be "unproofed" like the halting problem — Trestone
His new world is pure nonsense and fantasy for the Cave people. — Trestone
Show: There is no function from a set onto its power set.
Proof :
Let f be function from S to PS. Let d = {x | xeS & ~xef(x)}.
dePS.
If d is in range(f), then for some x in S we have d=f(x).
If xef(x), then ~xed, so ~xef(x).
If ~xef(x), then xed, so xef(x).
Contradiction. So d is not in the range of f. So f is not a function from S onto PS.
/
Show: ~ExAy yex.
Let Ay yex.
Let d = {x | xey & ~xex}.
If ded, then ~ded.
If ~ded, then ded.
Contradiction. So ~ExAy yex. — TonesInDeepFreeze
What is the first line in each of the below proofs that is not allowed in layer math?
Show: There is no function from a set onto its power set.
Proof :
Let f be function from S to PS. Let d = {x | xeS & ~xef(x)}.
If d is in range(f), then for some x in S we have d=f(x).
If xef(x), then ~xed, so ~xef(x).
If ~xef(x), then xed, so xef(x).
Contradiction. So d is not in the range of f. So f is not a function from S onto PS.
Note: In my previous posts, anywhere I mistakenly wrote 'level' I meant 'layer', as I guess would be obvious anyway.
↪Trestone
In ordinary mathematical logic, contradictions are syntactical, not requiring assignment of truth values. Meanwhile, as far as I can tell, your layer logic is described primarily semantically in terms of truth values; I don't know the syntax of whatever proof system you have in mind, so I can't evaluate the means by which you would prevent (syntactical) contradictions. You could assert that provability entails soundness, but we need to prove that, not just assert it, and you can't prove it without first stating what the proof system is. — TonesInDeepFreeze
Trestone: I do not fully understand, as I do not know the technical terms (syntax?, proof system?).
For me a contradiction is, if the same statement is shown as true and not true.
In layer logic the statement has to be in the same layer,
as being true in one layer and being false in another layer is allowed and no contradiction.
That stands without your response.
I would guess that layer logic does disprove contradictions. That is, layer logic disproves all formulas of the form 'P & ~P' (where they "reside" (or\e whatever way you say it) in the same level [should be layer]). — TonesInDeepFreeze
Trestone: No, only when there are different layers used.
In most classical proofs that are indirect or by contradiction,
different layers are used, if they are transferred to layer math,
so many are disproved, but not those in the same layer.
Is my guess correct?
And does layer math prove the following?:
~0=1 Trestone: false in layer math
and
~Ex (x is a natural number & x>x) Trestone: false in layer math
And you admit that layer math does not prove the fundamental theorem of arithmetic. So layer math would not seem to offer much as a mathematical foundation anyway.
Trestone: It is not so good for multiploikation and primes - b ut what if the real worl is so?
what you say is that there are three truth values and that statements are evaluated at different levels [should be layers]. You haven't given even the starting point: description of evaluation of truth and falsehood for atomic sentences, compound sentences, and quantificational sentences. — TonesInDeepFreeze
Trestone: In many cases it helps, that in layder 0 all sentences have truth value "undefined".
That often can be used for starting. More I have not looked upon yet.
That stands without your response.
how I handle the proof of the halting problem — Trestone
You begin with:
with layer logic we have to add layers if a program has to give a value/result:
A given program halts or not in layer k for given input data. — Trestone
But no axioms or rules of inference by which to claim that.
So to follow along with you in your layer math, one just has to accept the arbitrary lines in your arguments as given by you personally (there is no objective codification). You do not provide one with a way to check whether the lines you put forth are axioms or theorems of layer math but instead one must rely solely on your dicta as to what constitutes a valid line or inference in an argument.
Trestone: Yes, I have not developed a full layer informatics, I just added layers to programms,
that give a result. That was enough to abandon the Halting problem and
create Non-Turing algorithms.
my earlier handling of Cantor´s diagonalization and proof in layer logic — Trestone
You begin with:
(t marks the layers, W(x,t) ist the truth value of x in layer t, -w stands for „not true“ or „false“
ther value „undefined“ I left out to make things easier).
Be M a set and P(M) its power set and F: M -> P(M) a bijection between them (in layer d)
Then the set A is defined with W(x e A, t+1) = w := if ( W(x e M,t)=w and W(x e F(x),t)=-w ) — Trestone
In ordinary logic, truth values apply to sentences. It seems that had previously been the case in your discussion of layer logic too. Here you mention the truth value of x, So I take it that x ranges over sentences there. But then we find x ranging over prospective members of the set A. So which is it? x ranges over sentences or x ranges over prospective members of sets? So far, what you've given is pseudo-math or gibberish dressed up with undefined math/logic-sounding verbiage.
Trestone: x is a member of a layer set and therefore itself a layer set.
Also, you mention things (which I guess are sentence) as being true or false in layers, but now here we find that functions too are things in layers. But you've not stated what a layer sis or what kinds of things can be in layers or, as I mentioned earlier, how it is determined a given atomic, compound, or quantificational sentence is true or false in a layer.
Trestone: Yes, I am not very precise. Everything where you can ask if it has a truth value
(is it true, falser or undefined?) needs a layer in layer logic/math.
F: M -> P(M) a bijection — Trestone
Are you there asserting that there exists such an F? If you are, but without first proving the existence of such an F, it would seem to be question begging, since by supposedly refuting Cantor's theorem, you're claiming to prove that there does exist such an F.
Trestone: like in the proof of Cantor, I asume herre that such a F exists.
the proof about the power set can be similary be "unproofed" like the halting problem — Trestone
Just to be clear, these are all distinct:
(1) A proof of ~P in a given system..
(2) A meta-proof that P is not a theorem of given system.
(3) Pointing out a line in a purported proof of a given system that it is not actually an allowed line in that system (i.e. pointing out where a purported proof is not an actual proof).
(4) A meta-proof that P is false in a given model of a given theory.
So, letting P = Cantor's theorem, do you you claim either (1) or (2) regarding layer math? (I take it that you do claim (4) or something like it.)
Trestone: (5) I do not disprove the original P of Cantor,
but a transferred P2 in a new model, layer math.
His new world is pure nonsense and fantasy for the Cave people. — Trestone
That's question begging. One can just as well say you've not left your own cave, as you are not familiar with the logic and mathematics that has been explored by generations of logicians and mathematicians who have themselves studied alternatives including types, orders, levels in set theory, quantification over theories themselves, modalities, possible world semantics, topological semantics, and even para-consistency. — TonesInDeepFreeze
I am sorry that I can not answer most your questions to formal details. — Trestone
F: M -> P(M) a bijection between them — Trestone
in layer math, the existence of F does not lead to a contradiction — Trestone
hope you have a little understanding for a "Columbus" — Trestone
I do not know the technical terms (syntax?, proof system?). — Trestone
For me a contradiction is, if the same statement is shown as true and not true. — Trestone
indirect or by contradiction — Trestone
That is, layer logic disproves all formulas of the form 'P & ~P' [?'] — TonesInDeepFreeze
No, only when there are different layers used. — Trestone
~0=1 Trestone: false in layer math — TonesInDeepFreeze
~Ex (x is a natural number & x>x) Trestone: false in layer math — TonesInDeepFreeze
In many cases it helps, that in layder 0 all sentences have truth value "undefined". — Trestone
Non-Turing algorithms — Trestone
You begin with:
(t marks the layers, W(x,t) ist the truth value of x in layer t, -w stands for „not true“ or „false“
ther value „undefined“ I left out to make things easier).
Be M a set and P(M) its power set and F: M -> P(M) a bijection between them (in layer d)
Then the set A is defined with W(x e A, t+1) = w := if ( W(x e M,t)=w and W(x e F(x),t)=-w ) — TonesInDeepFreeze
like in the proof of Cantor, I asume herre that such a F exists. — Trestone
x is a member of a layer set and therefore itself a layer set. — Trestone
the proof about the power set can be similary be "unproofed" — TonesInDeepFreeze
"You will know them by their fruits" (Matthew 7:15-20) — Trestone
~0=1 Trestone: true in layer math — Trestone
~Ex (x is a natural number & x>x) Trestone: true in layer math — Trestone
On one day we get for a number n the prime decomposition P1.
One week later we get on the same computer with the same program for n
another prime decomposition P2 (and similar disturbing results with other computers). — Trestone
now I am used to being “a voice crying in the wilderness”. — Trestone
maybe unconsciously I want to be the only one
who understands Layer Logic, — Trestone
Logic/math statements do not refer to any event (real or hypothetical) in the physical universe, but are only true or false depending on the rules within the particular mathematical/logical system framework being used. — EricH
The term "classical logic" is a bit vague, — EricH
Perhaps there is a way to translate into classical logic syntax (it's beyond my capabilities) but I'm reasonably confident that even if the sentence could be formulated it would have a value of false — EricH
Put differently, the sentence "This sentence is false" does not express a coherent thought — EricH
even if the sentence could be formulated it would have a value of false — EricH
On the other hand I am interested to learn,more about the liar, extensions and the solutions? — Trestone
The term "classical logic" is a bit vague, — EricH
Classical logic is exactly formalized. It's not vague. — TonesInDeepFreeze
the OP does not make much sense — EricH
the SEP article on Classical Logic Is this your understanding of the term classical logic? — EricH
you remind me of the border guard
who demanded the TAO-TE-KING from Laotse. — Trestone
Our logic is not only two thousand years old,
it is also the basis of all science and only those who are stupid
and not suitable for true science cannot understand it, because it is very easy. — Trestone
the arithmetic showed
that most true sentences could not be proven — Trestone
They did not want to show any nakedness and emphasized the universal validity and unquestionable truth of the new logic. — Trestone
it doesn't work! — Trestone
The logic of Troy — Trestone
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