• Trestone
    51
    Hello,

    the liars sentence is a famous antinomy / paradox of classical logic:

    L:= „This statement is not true“

    If we assume, that L is true, then because of the definition of L it has to be not true (=false).
    If we assume, that L is not true, then because of the definition of L it has to true
    So in both cases L is true and false at the same time, and that is not allowed.

    Most solutions of the problem do not allow sentences like L and especially no selfreferences.

    But there is an other posiibility: A new logic called „layer logic“ or should I say „liar logic“?
    The idea with layer logic is, that every determination of a truth value
    belongs to a layer (0,1,2,3,...),
    and different truth values are allowed in different layers (whatever layers are – see beyond).

    As layers are „blind“ to themselfes and to higher layers,
    we always have to use a higher layer if we „talk“ about a layer.

    In the rules of layer logic we have this formulation of the liar L:

    „For all k=0,1,2,3,...: This proposition LL is true in layer k+1, if LL is not true in layer k
    and LL is false in layer k+1 else.“

    In layer logic all propositions are „undefined“ in the lowest layer 0.
    Therefore LL is undefined in layer 0.

    Layer 1: This proposition LL is true in layer 0+1, if LL is not true in layer 0
    and LL is false in layer 0+1 else..“
    Therefore LL is true in layer 1.
    Layer 2: This proposition LL is true in layer 1+1, if LL is not true in layer 1
    and LL is false in layer 1+1 else..“
    Therefore LL is false in layer 2.
    Layer 3: This proposition LL is true in layer 2+1, if LL is not true in layer 2
    and LL is false in layer 2+1 else..“
    Therefore LL is true in layer 3.
    Layer 4: This proposition LL is true in layer 3+1, if LL is not true in layer 3
    and LL is false in layer 3+1 else..“
    Therefore LL is false in layer 4.
    We see that LL has an alternating truth value with increasing layers – there is no contradiction.

    As propositions belong to all layers, LL is self-referential but not within a layer.

    And what is with liars that speak about all layers?
    „LA:= This proposition is not true in all layers“
    In layer logic meta propositions about layers and truth values have to be nearly classic:
    They can only be true or wrong and have to have the same truth value in all layers >=1.
    So if LA is true in layer 1 it has to be true in all layers, so LA is false – in all layers, that would be a contradiction.
    If LA is not true in layer 1 it has to be not true in all layers, so LA is true – in all layers, that would be a contradiction.
    Therefore LA is not an allowed meta proposition in layer logic.

    Maybe there are better examples for extended liars?

    The layers may look somehow strange at first glance.
    The layers were first just a formal parameter to differentiate truth values
    (for example in the first part of a proof and the second).

    Meanwhile I see them as a kind of a new dimension, of meta levels or cause and effect order
    or a new part of time.

    Even without knowing exactly what a layer is, we can use layer logic and layer set theorie,
    as the rules for using them are mostly independent of this.

    In a restricted way Professor Ulrich Blau in Munich invented a logic with layers
    some 20 years before me, the reflection logic.
    He counted how often we reflected about the truth value of a proposition
    and those meta levels were his layers.

    Here links with more detailed information about layer logic and layer set theory:
    layer logic on researchgate

    layer logic on The Philosophy Forum

    In German:
    www.ask1.org/threads/stufenlogik-trestone-reloaded-vortrag-apc.17951/#post-492741

    About Professor Ulrich Blau:
    https://ivv5hpp.uni-muenster.de/u/rds/blau_review.pdf

    In German:
    https://link.springer.com/chapter/10.1007/978-94-017-1456-3_20
    https://books.google.de/books?id=9xIxX206r5IC&pg=PA113&lpg=PA113&dq=reflexionslogik+blau&source=bl&ots=j5toa2ZhRK&sig=QWOej44nO1Upckoc9lEyfuZ1UEc&hl=de&sa=X&ved=2ahUKEwih-4Hm0erfAhVLaVAKHTwWDCsQ6AEwAXoECAkQAQ#v=onepage&q=reflexionslogik%20blau&f=false

    As it is unusual and bulky I can understand that not many are going to study layer logic,
    but in my eyes the possible results – a new look on logic and the world
    and a way out of many antinomies - it is worth the effort.

    On the other hand I am interested to learn,more about the liar, extensions and the solutions?

    Yours
    Trestone
  • keystone
    17
    I haven't read your work, but how would you handle the following pair of statements?
    Statement A: Statement B is true.
    Statement B: Statement A is false.
  • Trestone
    51
    Hello keystone,

    here the definitions (AL and BL) for statements A and B in layer logic:
    For all k=0, 1,2,3, …: Statement AL is true in layer k+1 if statement BL is true in layer k
    and statement AL is false else.
    Statement BL is true in layer k+1 if statement AL is not true in layer k
    and statement BL is false else.

    In layer 0 both AL and BL are „undefined“ (as always in layer 0).

    Layer 1: Statement AL is true in layer 0+1 if statement BL is true in layer 0
    and statement AL is false else.
    Statement BL is true in layer 0+1 if statement AL is not true in layer 0
    and statement BL is false else.
    Therefore statement AL is false in layer 1 and BL is true in layer 1.

    Layer 2: Statement AL is true in layer 1+1 if statement BL is true in layer 1
    and statement AL is false else.
    Statement BL is true in layer 1+1 if statement AL is not true in layer 1
    and statement BL is false else.
    Therefore statement AL is true in layer 2 and BL is false in layer 2.

    So AL and BL have alternating truth values in the layers and are no problem.
    As shown layer logic can also handle antinomies without direct selfreference.

    Yours
    Trestone
  • keystone
    17
    I see. With layer logic there is no self reference since layers cannot talk about themselves. The problem is that for all statements you've abandoned simple truth values and replaced them with infinite arrays of truth values. Does 1+1=2? Your answer would be [U,T,T,T,T,...]. Aside from being a bloated framework, your view rests upon the existence of actual infinities, 'numbers' which I have hesitations about.
  • Trestone
    51
    Hello keystone,

    yes, I use my layers for all statements and their truth values.
    For the layers I do not really need „numbers“, I only need a set that is inductive,
    multiplication is not needed for the layers.

    As the proof for Cantor´s diagonalization is valid no more with layer logic,
    one kind of infinity (that of the natural numbers) is enough.

    But the situation for mathematics is not all good:
    As the proof for the uniqueness of the prime factorization is no more valid,
    there might be different factorizations in different layers.
    This could show a way to proof or falsicate layer logic experimentally,
    but the needed numbers could be astronomically large.

    Layer logic is selfrefertial as the truth value of statements can be defined with the help
    of truth values of this statements.
    But layer logic is not fully self referential, as statements are not allowed to use truth values
    of the same (or higher) layers.
    The exciting question is, if this layer selfreference is needed.
    With layer logic I could define a set theory and natural numbers with arithmetics
    that followed the rules of layer logic.

    So the the approach carries further than I thought in the beginning ...

    Yours
    Trestone
  • jgill
    1.2k
    But the situation for mathematics is not all good:
    As the proof for the uniqueness of the prime factorization is no more valid,
    there might be different factorizations in different layers.
    Trestone

    At first glance I might think, How absurd. But these days there are lots of things in highly abstract mathematics that are beyond my pale, so maybe there is something to your ideas. I'm curious what fishfry and fdrake might think of it. In a faint way it resembles Schrödinger's cat. Nice article on Blau.
  • keystone
    17
    And what is with liars that speak about all layers?
    „LA:= This proposition is not true in all layers“
    In layer logic meta propositions about layers and truth values have to be nearly classic:
    They can only be true or wrong and have to have the same truth value in all layers >=1.
    So if LA is true in layer 1 it has to be true in all layers, so LA is false – in all layers, that would be a contradiction.
    If LA is not true in layer 1 it has to be not true in all layers, so LA is true – in all layers, that would be a contradiction.
    Therefore LA is not an allowed meta proposition in layer logic.
    Trestone

    If you are fine with disallowing statements in Layer Logic for which a truth value cannot be assigned, why not skip this added layer of complexity and disallow the original liar's statement in Classical Logic?
  • Trestone
    51
    Hello keystone,

    in Layer Logic I have to “disallow” only some meta statements.

    If I would disallow the original liar's statement (and renounce layer logic)
    I would loose a lot of new solutions and possibilities that come with Layer Logic:

    Besides the liar for example Cantors Diagonalization, Russell`s Paradox,
    probably also the Uncompleteness Sentences of Gödel
    and the Halting Problem of Informatics.
    And there are new possibilities for the philosophy of mind and probably for physics.

    So the „little effort with the layers“ should be worth the effort ...

    Yours Trestone
  • Trestone
    51
    Hello jgill,

    Thank you for your comment.

    To „Schrödinger`s cat“:
    I have tried to apply layer logic on Quantum theory (for double slit and entanglement).
    I constructed a nice model with inverse time for virtual particels
    (because they move in an „invisible“ layer),
    but as I am not a physicist it is mostly speculation.

    Yours
    Trestone
  • keystone
    17
    @Trestone

    So my understanding is that your resolution to the Liar's Paradox is not Layer Logic, but disallowing problematic statements.

    As for Russell's Paradox, isn't that what is basically done with ZF, where we 'disallow' the set of all sets that are not members of themselves? With this approach Russell's Paradox has been resolved and there's no need for Layer Logic.

    Personally I think this sort of disallowing problematic objects sweeps the problems under the rug.
  • Trestone
    51
    Hello keystone,

    my motivation for developing Layer Logic was to look after the things „under the rug“.
    Layer set theory is rather nice: It has only one kind of infinity, the Russell set
    and the set of all sets are ordinary sets.
    Layer Logic and layer set theory are not nescessary, but they offer a new perspective
    for the things „under the rug“.

    Yours
    Trestone
  • keystone
    17
    But in a way you are sweeping "this proposition is not true in all layers" under the rug. I wonder if there is an analog to the Russell set which you are also sweeping under the rug...
  • TonesInDeepFreeze
    264
    Just to be clear, the existence of a set whose members are all and only those sets that are not members of themselves is ruled out by first order logic alone, even before we get to set theory.

    For any two-place predicate F (whether F is the membership relation of any other relation) we have a theorem of first order logic:

    ~ExAy(Fyx <-> ~Fyy)

    So set theory does not need to add axioms that prove [read 'e' as epsilon]:

    ~ExAy(yex <-> ~yey)

    What set theory does need is to be careful not to add axioms that prove:

    ExAy(yex <-> ~yey)
  • Trestone
    51
    Hello,

    here some points about layer set theory and the Russell set,
    so you can check for yourself what is being swept „under the rug“:

    S1: Definition of layer sets
    The (layer set) x is in layer t+1 an element of the (layer) set M if and only if
    x has the property P(x) in layer t.
    In formulas:
    The truth value of „x e M“ in layer t+1
    is the truth value of P(x) in layer t.
    Vt >=0: Vx VM: W(x e M,t+1):= W(P(x),t)

    S2: Layer sets to propositions:
    For every layer logic proposition P(x) about any layer set x
    there exists a layer set M which fulfills for all layers t=0,1,2,3,...
    the element equation:
    W(x e M, t+1) := W(P(x), t)
    That means:
    The truth value of „x e M“ in layer t+1 is the same as
    the truth value of P(x) in layer t.

    In layer set theory „x e M“ can have three values: true, false and undefined.

    S3: Empty layer set:
    For the empty layer set 0 every set x is a Not-element,
    that means „x e 0“ has the truh value false in all layers >0.
    Vt>0: W(x e 0, t) := false and W(x e 0, 0) := undefined.

    S4: The full layer set All:
    For the full set All every set is an element in all layers >0.
    Vt>0: W(x e All, t) := true and W(x e All, 0) := undefined.

    Russel set with layers:
    We define the Russell layer set R as follows:
    x is element of R in layer t+1 if x is not element of x in layer t
    and not an element in layer t+1 else.

    Now we look on „R e R“ in different layers:

    In layer 0 „R e R“ is undefined (as always in layer 0 in layer logic).
    As undefined, R is not element of R in layer 0.
    Therefore „R e R“ is true in layer 1 (=0+1).
    Therefore „R e R“ is false in layer 2 (=1+1, not an element becaus of the „else“)
    Therefore „R e R“ is true in layer 3.
    Therefore „R e R“ is false in layer 4. And so on.

    As alternating truth values in the layers are allowed in layer set theory,
    the layer Russell set is an allowed set and not paradox.

    Yours
    Trestone
  • TonesInDeepFreeze
    264
    What are the primitives of your system? What are the formation rules?
  • Trestone
    51
    Hello TonesinDeepFreeze,

    as I left university some 30 years ago and have developed layer logic only as a hobby,
    I have no systematic theory, I mostly developed it in (German) discussions in threads like this.

    Most details in English can be found here:
    Layer logic at researchgate

    Some more details in German here:
    Layer Logic in German at ask1.org

    I would be glad if someone would start a systematic work on Layer Logic.

    Yours
    Trestone
  • maytham naei
    16


    I had created a thread with a similar topic
    https://thephilosophyforum.com/discussion/10641/how-the-greatest-lies-contain-the-greatest-truths

    I am going to copy the relevant part. It would be interesting to hear your thoughts on it.

    Let's start with a logical paradox:

    > "I always lie"

    Here is the problem with this statement:

    - If that statement is a lie: then I lied about "always lying", which means I must have told the truth at some point.
    - If that statement is the true: then I don't always lie, because I just told the truth.

    Thus one can never say "I always lie"

    Now let's add a bit of truth to the first statement.

    > "I mostly lie"

    Hopefully you don't know people in your life who mostly lie. But it's still possible for someone to do so.
    maytham naei
  • TonesInDeepFreeze
    264


    I can't make sense of your essay - from the very start where you begin by flinging around terminology used in a personal way but that you don't define nor declare as primitive.

    If you are truly interested in people spending their valuable time to grasp your ideas, then you would be well served by formulating them systematically rather than merely hoping that Internet passerbys would try to figure out the meaning of your thrown-together verbiage for you.
  • TonesInDeepFreeze
    264
    Thus one can never say "I always lie"maytham naei

    One can say 'Everything I say is false'.

    But one can't say it without self-contradiction.

    (By the way, for purposes of paradox, it's clearer to refer to 'truth' and 'falsehood' rather than 'lying' and 'not lying'. Because someone can state a falsehood with lying (the person may not know that their statement is false).

    In particular, you conflated lying with not telling the truth when you made the alternatives:

    Lied vs told the truth.

    The proper alternatives are either

    Lied vs did not lie

    or

    Told truth vs didn't tell truth.

    > "I mostly lie"

    Hopefully you don't know people in your life who mostly lie. But it's still possible for someone to do so.
    maytham naei

    Yes, 'I mostly lie' is not self-contradictory. So what?

    Hopefully you don't know people in your life who mostly lie.maytham naei

    For four years the people of the United States had the 45th president in their lives who lied as many times a day as most people breathe in and out. Probably he even lied in his sleep.
  • Trestone
    51
    Hello maytham naei,

    I think the logic problems with „I always lie“ are similar to „This statement is not true“.
    With classic logic they are true and false – what is not allowed.

    In layer logic we would formulate for example:

    LA is true in layer k+1 if LA:= „All I say“ is not true in layer k and LA is false else.

    In layer 0 LA is undefined (as all layer logic propositions).

    In layer 1:
    LA is true in layer 0+1 if LA:= „All I say“ is not true in layer 0 and LA is false else.
    Therefore LA is true in layer 1.

    In layer 2:
    LA is true in layer 1+1 if LA:= „All I say“ is not true in layer 1 and LA is false else.
    Therefore LA is false in layer 2.

    So “ I always lie“ or „All I say is not true“ in layer logic are allowed statements
    that have alternating truth values „true“ and „false in the layers >0.

    Yours
    Trestone
  • Trestone
    51
    Hello TonesInDeepFreeze,

    Lenin is quoted with the following syaing:

    “If these Germans want to storm a train station, they first buy a platform ticket!”

    I am myself feeling like a German revolutionary / explorer.
    But not so much as a scientist but more like Christopher Columbus:
    I detected Layer Logic by chance as I am more on a philosophical than a logical journey.

    My goal is to pass on the revolutionary intuition associated with the Layer Logic.
    And the most important points thereby are not “platform tickets” or even Layer Logic,
    but to open new ways of thinking besides classic logic,

    Layer Logic itself to me looks not so complicated to understand,
    but here some more explanations that might help:

    The only new components are the layers.
    Eight things are important about layers:

    A) The layers are elements of an inductive set with elements 0,1,2,3, …
    (multiplicative properties not needed)

    B) All propositions P have truth values W only in combination with a layer k: W(P,k).

    C) There are 3 possible truth values W(P,k): true (=t) , false (=f) and undefined (=u)

    D) In layer 0 all Propositions P are undefined: For all P: W(P,0)=u

    E) The truth value of a proposition P is the vector of all the truth values in all layers.
    W(P) = (u, W(P,1), W(P,2), W(P,3), …)

    F) A proposition P is well defined, if the truth values W(P,k) for all layers k
    are well defined (one value for every layer).

    G) When defining values for W(P,k+1) for proposition P all defined propostions and values
    of smaller values (k or smaller) can be used - even W(P,k).

    H) Layers and Propositions in layer k are „blind“ for this layer k and higher layers.
    So when speaking about a property or using a value we have to change from layer k to k+1.

    Analysis of most classical indirect proofs show, that with layer logic we have because of G) and H)
    to use two different layers.
    As true and false in different layers is allowed in layer logic there is no more
    a contradiction and the indirect proofs are valid no more.
    (Within a layer different truth values are still not allowed).

    That is the revolutionary part of Layer Logic!

    With all this formalization we still do not know what layers are
    and why we did not notice them (or the new dimension) in the last 2000 years?

    Well, I have already used and showed layers with the liar and with Russell`s set.
    In everyday use most propably layers make no difference,
    as properties may change with layers but they do not have to.

    So layers mostly make a difference with infinity, selfreference and the start of cause - effect chains.

    And in everyday life we all can be in the same layer that may change (simultanously)
    with every physical interaction (besides gravitation) – but that is very speculative.
    So that if two people look at an objekt at the same time,
    they see the same propositions in the same layer.

    But may be the main reason why we do not perceive layers could be,
    that they don't fit into our view of the world ...

    Prof. Ulrich Blau gave a more formerly definition of his reflexion logic
    and his layers as „levels of reflection“ -
    and he wrote a (German) book with about 1000 pages around it.

    About Professor Ulrich Blau:
    review about Prof. U. Blau

    In German:
    German Link 1 about Prof. Blau Reflexionslogik

    German link 2 on Prof. Blau Reflexionslogik

    His reflexion logic is only for a small part of all propositions, the reflective propositions,
    where as Layer Logic treats all propositions in the new way –
    as a full new logic with a new dimension, the layers.

    By the way:
    When I went to my first demonstration in 1989 UniMut in Berlin,
    I actually bought a subway ticket before to get to the KuDamm,
    where our students demonststration took place.
    A revolutionary student theatre had agitated me.

    Later I voted in Marburg against student strikes, but took a prominent part later.
    With philosophy students and professors we performed a play of me
    (“The death of Sokrates”) with also contained a (Sophistic) saying about logic:

    “If logic does not apply, it can confidently continue to apply -
    and that is also still thought of logically!”

    So you see, I've been dealing with platform cards and logic for 30 years.

    Unfortunately, my creativity and intuitions are dwindling
    so I have to talk about my ideas from 30 years ago ...


    Yours
    Trestone
  • keystone
    17
    @Trestone
    I haven't had the time to read through your messages thoroughly so I don't expect a response if you've already addressed my comment in your earlier posts.

    You disallowed "This proposition is not true in all layers" in Layer Logic. Is there something analogous in Layer Set Theory, perhaps something like "x is an element of R if it is not an element of R in all layers"?
  • TonesInDeepFreeze
    264
    Analysis of most classical indirect proofs showTrestone

    If you have in mind the famous proofs regarding a universal set, uncountablity, incompleteness, Tarski's theorem, and the halting problem, then these have direct proofs. Any proof of the form.

    Show ~P.
    Assume P.
    Derive contradiction.
    Conclude ~P.

    has a structure of direct proof.

    Indirect proof is of the structure:

    Show P.
    Assume ~P
    Derive contradiction.
    Conclude P.

    The non-existence of a universal set, uncountability, incompleteness, Tarski's theorem, and the halting problem do not rely on that structure.
  • Trestone
    51
    Hello keystone,

    the statement "x is an element of R if it is not an element of R in all layers"?
    is not allowed as a layer theory statement, as there is no layer given
    where "x is an element of R" should be true.
    if we add an layer k, the "all layers" will break the rule, that in definitions only
    smaller layers are allowed.
    So the forbidding of statements is according to rules.

    Yours
    Trestone
  • Trestone
    51
    Hello TonesInDeepFreeze,

    perhaps the formulation „indirect proof“ was misleading.
    I better could say „proof by contradiction“.

    The point is, however we call those proofs, the constructed contradiction in them
    is not valid any more when we transfer the proofs to layer logic.

    The reason is, that by constructing the contradictions we have to use different layers,
    and different truth values in different layer are not a contradiction in layer logic.

    And yes, all the proofs you named are valid no more
    and probably also the uncompleteness sentences of Gödel
    (I did not proof this completely with layer logic so far).

    Yours
    Trestone
  • TonesInDeepFreeze
    264
    No, neither 'indirect proof' nor 'proof by contradiction' are correctly applied to those proofs, as they are not of the form:

    Show P.
    Assume ~P
    Derive contradiction.
    Conclude P.

    by constructing the contradictions we have to use different layers,
    and different truth values in different layer are not a contradiction in layer logic.
    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
    264
    all the proofs you named are valid no moreTrestone

    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).

    /

    How does layer logic disallow this proof (which is not indirect)?:

    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.

    /

    How does layer logic disallow this proof (which is not indirect)?:

    Show: ~ExAy yex.

    Let Ay yex.

    Let d = {x | xey & ~xex}.

    If ded, then ~ded.

    If ~ded, then ded.

    Contradiction. So ~ExAy yex.

    /

    Remember, I never mentioned 'truth' or 'falsehood'. I merely gave syntactical proofs. So you haven't shown how those proofs are disallowed by merely saying that truth and falsehood may alternate on levels.

    As for semantics, basically, what you say is that there are three truth values and that statements are evaluated at different levels. You haven't given even the starting point: description of evaluation of truth and falsehood for atomic sentences, compound sentences, and quantificational sentences.
  • Trestone
    51
    Hello,

    here I will show how I handle the proof of the halting problem:
    Here a classical proof of it (from https://wiki.c2.com/?HaltingProblemDiscussions):

    "Assume I have a program P that can tell you whether any program halts or not for given input data.
    I construct a program Q based on P which, if P says its input program doesn't halt, immediately halts, and if P says the program halts, goes into an infinite loop.
    Feeding Q(Q) to P I can see that if P says it halts, it won't, and if P says it doesn't halt, it will.
    Therefore I don't have any such program P and anyone else who says they do is full of it."


    Now 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.
    And the program P has to tell about the halting in layer k+1,
    as only values oflower layers can be worked on.
    Now when constructing Q, which if P says in layer k+1 its input programm
    does not halt in layer k it immediately halts, that halt of Q will be in layer k+2,
    as values of layer k+1 are needed.
    And the infinite loop of Q in the other case will also be in layer k+2.
    We feed now Q(Q) to P :
    We look at Q in layer k. If P says in layer k+1 that Q halts on Q in layer k,
    the construction of Q says, that Q goes to an infinite loop in layer k+2.

    So Q halts in layer k and Q goes to an infinite loop in layer k+2.
    That is no contradiction, as programms can have different outcomes in different layers.
    (That is not the complete layer logic proof, but the main idea is given.)
    Therefore it is possible, that a Halting programm H exists,
    that gives a true in layer k+1 for every layer programm P,
    if P stops in layer k with input X.

    The layers make the difference.

    Yours
    Trestone
  • Trestone
    51
    Hello,

    the proof about the power set can be similary be "unproofed" like the halting problem
    by adding layers and layer logic.

    There is even a more simple "proof":
    In layer set theory the set of all sets (called All) is a set.
    The power set of All is All.
    So there is a bjection from a set (All) to its powerset (All) : the identity x->x.

    That is one of the things why I like the layer set theory.

    Yours
    Trestone
  • Trestone
    51
    Hello,

    I have found my earlier handling of Cantor´s diagonalization and proof in layer logic:

    As All, the set of all sets, is a set in layer theory, it is no surprise,
    that the diagonalization of Cantor is a problem no more (I just give the main idea):

     (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 )
    A is a subset of M and therefore in P(M).
    So it exists x0 e M with A=F(x0).

    First case: W(x0 e F(x0),t) = w , then W(x0 e A=F(x0), t+1) = -w
    (no contradiction, as in another layer)

    Second case: W(x0 e F(x0),t) = -w then W(x0 e A=F(x0), t+1) = w
    (no contradiction, as in another layer)

    If we have All as M and identity as Bijektion F we get for the set A:
    W(x e A, t+1) = w := if ( W(x e All,t)=w and W(x e x),t)=-w ) = if ( W(x e x),t)=-w )
    This is the layer Russell set R (I omitted the ´u´-value for simplification) -
    and no problem.
     (R is a regular set in layer set theory).

    So in layer theory we have just one kind of infinity – and no more Cantor´s paradise …

    A important remark: I do not say that the classic proofs are false, they are perfectly right.

    But with layer logic and layer set theory we are in a new world.
    All terms have to be transferred into the layer world
    and only there most of the proofs are valid no more.

    It is a little like in Plato's allegory of the cave:
    If someone has been out of the cave and seen the real world,
    he has to learn new rules and the rules of the old shadow world
    will no longer fit.
    If someone returns from the sun to the cave,
    nobody will listen to him or understand him.
    “Speak within our shadow rules or be quiet.”
    His new world is pure nonsense and fantasy for the Cave people.

    Yours
    Trestone
  • keystone
    17
    @Trestone

    My impression remains that you're adding on layers as a way to justify disallowing problematic statements/sets from existing within Layer Logic. My take is that if we wanted to avoid the problem by disallowing problematic statements we might as well do that within the simple framework of Classical Logic...but that's just my view. I hope you can find others to bounce ideas off of. If you can convince TonesInDeepFreeze that will be a good sign!
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