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Trestone

layer logic - alternative for humans and aliens?

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5 minutes ago, Trestone said:

About 15 years before me Ptrofessor Ulrich Blau in Munic had similar ideas, he called it “reflexion logic”.
For him layers were the times we reflected about a sentence (like the liar L “this sentence is not true”.
Layer 0: no reflection. L has the truth value “undefined”.
Layer 1: We reflect, that L was undefined in layer 0, therefore it is true.
Layer 2: We reflect on our reflection: L is false. Layer 3: L is true. And so on.

It has taken you 9 years to explain what your idea is. But thanks.

7 minutes ago, Trestone said:

that a computer that would use layer logic would not be limitid by the Halting Problem.

Can you show the proof of that?

I can guess there might be an alternative proof of the Halting Problem where you show that it requires an infinite number of layers. But I can't imagine how you could solve the halting problem. 

You have made all sorts of claims about what you can do, but so far these just seem be assertions (ie. with no support).

 

10 minutes ago, Trestone said:

so layer logic is more a philosophical theory

Logic in philosophy is a formal system, a branch of mathematics, so I would like to see some more support for your claims.

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Hello Strange,

my main point is, to imagine, that classical logic may be not the real logic for our world
and to construct an alternative.
I found layer logic as a good possiblity.

Of course it is very similar to classic logic (that sufficed for 2000 years),
and as the meta logic of layer logic I still use classical logic.

But with one parameter more (the layers), I can avoid almost all classical logical paradoxes
and most indirect proofs are valid no more, even as layer logic does allow indirect proofs
(within one layer).

Math and computing science are still possible,
but they are different in some points.

Here a short analysis of the holding problem from computer science and on "layer algorithms"
from the view of my “layer logic”:

In the layer logic, a new parameter is added to the programs, layer t.
A hierarchy applies:
If a program wants to evaluate / use a value from another program from layer t,
it can only do so at level t + 1 or higher (= t + r).

We are looking for a (layer) program H that decides on each program P with (string) input X in layer t + r,
whether this ever stops in step t or runs endlessly (e.g. due to a continuous loop).

Definition / basic property H (P, X, t + r):
The following applies to all programs P and inputs X:
IF P (X, t) stops THEN H (P, X, t + r): = true ELSE H (P, X, t + r): = false

In this case, r> = 1 must be selected, since layer t (at P) is used when calculating H.
More precisely, the next universal layer t + r (t) is to be set (see below).

We can now try to understand the classic counter-proof of the existence of H (P, X) and have to add the levels:

Suppose H (P, X, t + r) exists with the property required above.
(This is a hypothesis!)

Then we use H to construct a "strange" program S:

Definition S (P):
The following applies to all P:
S (P, t + r + k): = IF H (P, P, t + r) = true THEN loop ELSE S (P, t + r + k) = true; STOP
(In contrast to the meta / colloquial formulation at H, S (P) can be written as a real program if the code of H is available.
"loop" stands for a continuous loop)

Here k> = 1 should be selected, since the layer t + r (at H) is used when calculating S.
More precisely, the next universal layer t + r (t) + k (t + r (t)) has to be applied (see below).

S therefore uses the result of the holding program when applying a program P to its own source code as input.
Now we consider the self-application of S, i.e. we take the code for S as input for S

S (S, t + r + k) = IF H (S, S, t + r) = true THEN loop ELSE S (S, t + r + k) = true; STOP

Since H (S, S, t + r) = true exactly when S (S, t) stops, it is no longer paradoxical or contradictory:
S (S, t + k + r) loops when S (S, t) stops and stops when S (S, t) does not stop!

The following applies: t + k + r is not equal to t, i.e. two different layer calls from S.
So S is a program with different values at different layers, but not necessarily paradoxical.
S and H can therefore exist.

So there could be a holding program H with layer logic.

Why have I left r and k indefinite and not chosen 1 each?

Now I suspect that the layer of step programs does not only depend on the subroutines called
(they must be larger in each case),
but also of the interactions in the universe

(= the “layer of the universe " or at least of the “layer of the reference system”😞
 

My speculation:
Every interaction (except through gravitation) increases the layer counter in the universe (also in computers) simultaneously,
therefore the layers grow constantly and very quickly (and unfortunately hardly controllable).

The next processing layer for layer t can only be narrowed down
(at least 1 higher), but do not determine exactly.
It would be labeled t + r (t).

And we couldn't call computer programs twice with the same parameters ("don't go into the same river twice"),
because the universal step counter “flows” constantly, and we cannot enter the step t, but we find it again and again (and higher).
(Maybe it could be possible in the event horizon of a Black hole.)

If the layer logic applies, then today's computer programs probably only work because they are limited to layer-independent programs,
which is only a small part of the conceivable programs.

Despite the problems outlined, this does not have to stay that way
and maybe one or the other "layer" surprise is also possible in computer science ...

E.g. one could refute the "Curch-Turing thesis" with layer computers:
(i.e. calculate something new):

If you implement an algorithm P (X) on a (normal) computer,
the current layer t of the universe is implicitly used for each calculation,
i.e. the computer calculates P (X, t).
When P (X) is recalculated later, it calculates P (X, t + r).

If P (X, t) is a step-dependent function, P (x, t) and P (X, t + r) could be different,
although classically it should only be one value.

Such a function P (X) could be the prime number decomposition of X,
which could be layer-dependent for "large" X and t.
However, the numbers would probably have to be so large that a practical review in this way is not yet possible.

Maybe someone will try the experiment anyway or has an idea ...

Yours
Trestone

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1 hour ago, Trestone said:

The next processing layer for layer t can only be narrowed down
(at least 1 higher), but do not determine exactly.
It would be labeled t + r (t).

Your argument is pretty incoherent and your notation unclear but you have not prove that this "t+r(t)" does not potentially become infinite. (Which obviously is can do).

Therefore, you have not solved the halting problem.

Why you think black holes are relevant is beyond me. You don't seem to understand that mathematics is an abstract, formal system and has nothing to do with the universe or black holes (or even actual computers).

1 hour ago, Trestone said:

E.g. one could refute the "Curch-Turing thesis" with layer computers:
(i.e. calculate something new):

So we are back to unsupported claims.

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Hallo strange,

my proof is finished at the line:

“So there could be a holding program H with layer logic.”

Your quote " The next processing layer ...
It would be labeled t + r (t)."  is part of later speculations.

I did not solve the Halting problem but proofed,

that with layer logic the proof of the Halting problem is no longer valid.

In logical and mathematical layer logic I only use finite layers.
The infinite layer (with layer logic there is only one infinity) I use for philosophy and there as layer of the mind.

How nature and math are connected (for example pickets and natural numbers)
I do not know, but some kind of connection there seems to be.
And a logic that would have no connection to our reality/nature would be a strange and not very useful thing.

(Sorry for my unclear and clumsy notation, but my university time is past more than 30 years).


Yours
Trestone

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24 minutes ago, Trestone said:

“So there could be a holding program H with layer logic.”

This is only possible for a finite number of layers. The halting problem may require an infinite number and is therefore not solvable.

25 minutes ago, Trestone said:

I did not solve the Halting problem but proofed,

that with layer logic the proof of the Halting problem is no longer valid.

You may think that.

25 minutes ago, Trestone said:

In logical and mathematical layer logic I only use finite layers.

And you cannot solve the halting problem with finite layers. (This is simply equivalent to building a bigger or faster computer C+1 to find out if computer C will halt or not.)

 

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On 5/21/2020 at 5:46 PM, Trestone said:

I see no connection to the “layer model” in computing except the name.

Thank you for the response.

I did not say they were the same but I did (and still do) see some common features.

I was really trying to establish some point to your 'layer logic' as I cannot see a 'whole' to understand and appreciate.

There is a very clear reason for the 'layer model' and subsequent benefit from it, as exemplified in computing but which appears in many guises in most technical subjects.

If you don't wish to discuss that further, that's fine by me. I will leave you to your musing.

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Hello,

in my definitions for layer logic a statement does not belong to a layer,
but is independent of all layers.
The statements have a truth value in each layer (sometimmes different in layers)
and are mostly defined by recursion.

An example is the (layer) liar L:

For all t= 0,1,2, ...: The liar L is true in layer t+1 if it is not true in layer t – and false in layer t+1 else.

The layer truth vector of a layer statement is an infinite vector for t=0,1,2,3,... .

For the liar L it is (undefined, true, false, true, false, ...)

The same is with layer algorithems or programms P,
they are independent or comprehensive of layers -
and can stop in one layer t and not stop in an other t+1.

It is possible, that a Halting programm H exists,
that gives a true in layer t+1 for every layer programm P,
if P stops in layer t with input X.

(Important: The same H will give a true ore false for P and X for layer t+2).

So I think that there are not infinite Halting programms in layer logic.

(But of course I do not have H explicitely)

 

Yours
Trestone


 

Hello studiot,

in the TAO-TE-CHING (Lao Tzu) is the saying:
“A journey of a thousand miles begins with one step”.

May be layer logic and “layer model' in computing have this one step together.

But then the journey in my eyes goes different ways:

In layer logic I have infinite layers (0,1,2,3,...) and a strict hierarchy of layers:

In a lower (or equal) layer no information of a higher layer is accesible,
they are “blind” for all above.

And the truth values in the layers are recursivly used to define the truth values of a layer statement for all layers,
so layer statements are independent of layers (= defined for all layers).

And mainly of course it is a logic.

Layer logic has four fathers:

- Classic proposional logic (at least 65 % are the same)
- Three-valued logic (
Łukasiewicz logic) (using three truth values, 5%)
- The Logic of reflection by Prof. Ulrich Blau (using layers, only for reflecting proposals, 10%)
- Layer Logic of Trestone (expanding layers to all proposals, layer set theory, 20%)

It can be used very similar to classical logic:
For logic itself, for doing set theory and math, for computer science, for philosophy,
for physics and other science (or here for fiction).

Yours,
Trestone

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3 hours ago, Trestone said:

Hello studiot,

in the TAO-TE-CHING (Lao Tzu) is the saying:
“A journey of a thousand miles begins with one step”.

May be layer logic and “layer model' in computing have this one step together.

But then the journey in my eyes goes different ways:

In layer logic I have infinite layers (0,1,2,3,...) and a strict hierarchy of layers:

In a lower (or equal) layer no information of a higher layer is accesible,
they are “blind” for all above.

 

3 hours ago, Trestone said:

in my definitions for layer logic a statement does not belong to a layer,
but is independent of all layers.
The statements have a truth value in each layer (sometimmes different in layers)
and are mostly defined by recursion.

An example is the (layer) liar L:

For all t= 0,1,2, ...: The liar L is true in layer t+1 if it is not true in layer t – and false in layer t+1 else.

The layer truth vector of a layer statement is an infinite vector for t=0,1,2,3,... .

For the liar L it is (undefined, true, false, true, false, ...)

 

Thank you for your replies.

Unfortunately although they have shed some light, they leave me more puzzled than ever.

Both the quoted examples assert ( in different ways) the independence of one 'layer' from another.

So how is it possible to establish the 'truth value' of any succeeding or preceeding layer, given the truth value in a particular layer.

This must be so since with your simple example of the truth value of any layer immediately succeeding the given one must be opposite to the given value.

 

So there must be more to it than this

 

Since you like quoting Chinese philosophers, here is a problem that is solved in 12 'layers' posed sometime between 200 BC and 10 AD in the Ancient Chinese test

"Nine Chapters on the Art of Calculation"

In modern parlance it is

3x + 2y +z = 39
5y + z = 24
4y + 8 z = 39

 

Perhaps you would like to look at it from the points of view of your layers and my 12 layers, for comparison?

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Posted (edited)

I'm not entirely sure if this is relevant but we get quite a lot of "multiple truths" in law enforcement such as independent witnesses giving conflicting accounts of the same event, potentially false memories of childhood trauma, or the weapon-focus effect.

All these are influenced, in varying degrees, by an observers' unconscious bias, their (flawed?) perception or outside influences.

The knack it to test the observations and piece together the most likely "truth" then let a jury decide.  A bit like peer review, I suppose.

Edited by Dord

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13 minutes ago, Dord said:

I'm not entirely sure if this is relevant but we get quite a lot of "multiple truths" in law enforcement such as independent witnesses giving conflicting accounts of the same event, potentially false memories of childhood trauma, or the weapon-focus effect.

All these are influenced, in varying degrees, by an observers' unconscious bias, their (flawed?) perception or outside influences.

The knack it to test the observations and piece together the most likely "truth" then let a jury decide.  A bit like peer review, I suppose.

I'm not sure it's 'layer' relevant but it's a good point. +1

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Posted (edited)

Hello studiot,

it is ok to be puzzled by layer logic.

I myself was puzzled for years and still am in some questions.

When constructing propositions like the layer liar,
the connection between layers is easy and clear.

And the use of recursions helps to connect truth values of different layers.
(And layer logic helps with induction/recursion, as there is always a "free" start
with "all propositions have truth value undefined in layer 0")

But I am not sure if we have such connections between layers for all propositions.
If there are layers in the (logical) world, they could be also independent from each other.
So a layer could be like a world of its own.

There is a hierarchy with the layers, but this does not mean,
that a truth value in a higher layer is “more true” or “more important” than a truth value of a lower level.
All “exist” simultanously and equally.

As we all perceive a similar world and do seldom discuss
if there is an object or not (for example because of different properties in different layers)
we all seem to live in the same layer,
at least with our perception.

As layers increase with cause and effect, this layer is dynamic.
That was one of the reasons why I believed, that there is one layer for the whole universe,
and that it increases with every interaction (except gravitation) of objects in it.
I have learned to give a layer to every observer frame system.

But here we have left the reign of pure logic and changed to human perceiption and physics.

In your chinese problem I do not see the “12 layers”?

If a solution in natural numbers is looked for, the last line could be connected to layer logic:
4y + 8 z = 39 . Classical the left side is even, the right side not.
In the arithmetic of layer logic, a number can be even in one layer and not even in another,
so there could be a layer were there exists a solution in natural numbers.
(But I do not know numbers, that are even and not even in layers – and I have not solved the puzzle.)

All in all layer logic is in some respects similar to the “ Many World Theory”,
but I hope not too much, as I like this theory not at all.

Yours
Trestone

 

Hello Dord,

I think parts of classical logic were developed in ancient Greek for speaking at court and searching for truth.
As there is not “one truth” in layer logic (but one for each layer),
it could help to be more open minded and to tolerate even contradictory statements.

But I do not think that layer logic will help us in real live with witnesses at court:

I believe, that all humans are “in” the same layer when perceiving,
so the differences do not come by different layers but different people.

And humans are a greater mystery then logic...

Yours
Trestone

Edited by Trestone

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Posted (edited)

Hello,

I found a link on English to Prof. Ulrich Blau´s work  (numbers, paradoxes and relexion logic) (there are not many):

(It is formally more correct then my work, but really "hard stuff", most about regions of infinity that my layer logic no longer needs)

https://books.google.de/books?id=Xg6QpedPpcsC&pg=PA311&lpg=PA311&dq=Reflexion+logic+Ulrich+Blau&source=bl&ots=HZNNGnEARu&sig=ACfU3U2Hunv0FZTcdq9T5TDfVa7XArq9VA&hl=de&sa=X&ved=2ahUKEwjf9ZSVnczpAhVERxUIHXhHAp44ChDoATALegQICBAB#v=onepage&q=Reflexion%20logic%20Ulrich%20Blau&f=false

Yours
Trestone

Edited by Trestone

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25 minutes ago, Trestone said:

I found a link on English to Prof. Ulrich Blau´s work  (numbers, paradoxes and relexion logic) (there are not many):

Thank you for this and your previous reply.

 

However I was really looking for specifics wbout your system we could discuss but you made no refernce to it.

23 hours ago, studiot said:

Thank you for your replies.

Unfortunately although they have shed some light, they leave me more puzzled than ever.

Both the quoted examples assert ( in different ways) the independence of one 'layer' from another.

So how is it possible to establish the 'truth value' of any succeeding or preceeding layer, given the truth value in a particular layer.

This must be so since with your simple example of the truth value of any layer immediately succeeding the given one must be opposite to the given value.

 

As to my chinese problem, I was suggesting running the solution  in your layer logic scenario and what I understand by layers for comparison purposes.

But you have not doen this you you now assert that your method cannot solve those three simultaneous equations?

Note the chinese wrapped the problem up in many words; I have just posted the core math of it in modern algebra.

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Hello studiot,

I do not know all about layer logic and layer arithmetics,
as I just made some definitions to have a layer atrithmetic - but I have no practice in using it.

I looked if with layer logic a solution for your Chinese problem would be possible with natural numbers
(and different layers) - but I do not know enough of layer arithmetics.

Within one layer the solution is the same as in classical arithmetic and we get three integer fractions as solutions,
but we learn nothing about layers this way, as there are the same rules for arithmetics within one layer as in classical arithmetic.

Yours
Trestone

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6 minutes ago, Trestone said:

Hello studiot,

I do not know all about layer logic and layer arithmetics,
as I just made some definitions to have a layer atrithmetic - but I have no practice in using it.

I looked if with layer logic a solution for your Chinese problem would be possible with natural numbers
(and different layers) - but I do not know enough of layer arithmetics.

Within one layer the solution is the same as in classical arithmetic and we get three integer fractions as solutions,
but we learn nothing about layers this way, as there are the same rules for arithmetics within one layer as in classical arithmetic.

Yours
Trestone

 

Again thank you for your response.

It is most interesting that you think to know all about my version of layers (especially when I haven't yet posted them) but fail to answer when I point out a contrdiction in the implementation of yours.

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