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  • The Paradox of Symmetric Induction

    From William Elliot@21:1/5 to All on Mon Mar 6 09:36:21 2017
    Let K be a nonempty set and S:K -> K a function.

    Restrict lower case variables to K
    and upper case variables to P(K).

    Assume biduction, also known as symmetric induction:
    for all nonempty A,
    if for all x in A, (x in A iff Sx in A)
    then A = K.

    Algelo Margaris, "Successor Axioms for the Integers"
    uses symmetric induction to axiomatically describe the
    integers.

    (K,S) with biduction is a consistent theory.
    For example the integers, the positive integers,
    or the integers modulus n with Sx = x + 1 are
    models that satisfy the axioms.  K could be finite
    and (K,S) would still be a consistent theory.

    Use biduction to define a relation < with
    not a < a;  a <= b iff a < Sb
    where a <= b is written for a < b or a = b.

    Here's my reasoning why 'a <' is defined for all of K.
    First off a < a is defined as false.
    Whenever a < b is defined, a < Sb is defined as a <= b
    Whenever a < Sb is defined, a < b is defined as a < Sb and a = b.
    This later come from the definition since
    a <= b iff a < Sb
    is equivalent to
    a < b iff a < Sb, a /= b.
    Thus using biduction conclusion 'a <' is defined for all of K,

    However with the establishment of that definition the previous
    finite models create contradictions.  Nothing so simple as < is
    empty or < equals KxK.  No, a raw a < a.

    For example, K = {0,1,2}, S0 = 1;  S1 = 2;  S2 = 0.
    0 <= 0;  0 < S0 = 1;  0 <= 1
    0 < S1 = 2;  0 <= 2;  0 < S2 = 0

    What's wrong with the definition?  
    Where is the blunder in the above reasoning?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From David Hobby@21:1/5 to William Elliot on Tue Mar 7 12:02:35 2017
    On Monday, March 6, 2017 at 11:36:26 AM UTC-5, William Elliot wrote:
    ...
    Assume biduction, also known as symmetric induction:
    for all nonempty A,
      if for all x in A, (x in A iff Sx in A)
    then A = K.

    For an arbitrary function S, you only get that A is closed under image
    and inverse image.  Yes, K is such a set, but subsets of K may be as
    well.
    ...
    Use biduction to define a relation < with
      not a < a;  a <= b iff a < Sb
    where a <= b is written for a < b or a = b.

    Here's my reasoning why 'a <' is defined for all of K.
    First off a < a is defined as false.
    Whenever a < b is defined, a < Sb is defined as a <= b
    Whenever a < Sb is defined, a < b is defined as a < Sb and a = b.

    Assuming that S is a function where this is valid, the previous two
    lines define the predicate P(x) = "a < x" for all x in K.  But there's
    no reason that P(x) defined this way has "not P(a)".  You've added a <
    a as an extra condition; there's no reason that this needs to be
    consistent with the rest of your definition of <.
    ...
    However with the establishment of that definition the previous
    finite models create contradictions.  Nothing so simple as < is
    empty or < equals KxK.  No, a raw a < a.

    For example, K = {0,1,2}, S0 = 1;  S1 = 2;  S2 = 0.
    0 <= 0;  0 < S0 = 1;  0 <= 1
    0 < S1 = 2;  0 <= 2;  0 < S2 = 0

    What's wrong with the definition?  
    Where is the blunder in the above reasoning?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From William Elliot@21:1/5 to David Hobby on Sun Mar 12 13:27:54 2017
    In article <070320171202351509%edgar@math.ohio-state.edu.invalid>,
    David Hobby <david.c.hobby@gmail.com> wrote:

    On Monday, March 6, 2017 at 11:36:26 AM UTC-5, William Elliot wrote:
    ...
    Assume biduction, also known as symmetric induction:
    for all nonempty A,
      if for all x in A, (x in A iff Sx in A)
    then A = K.

    For an arbitrary function S, you only get that A is closed under image
    and inverse image.  Yes, K is such a set, but subsets of K may be as
    well.
    ...
    Use biduction to define a relation < with
      not a < a;  a <= b iff a < Sb
    where a <= b is written for a < b or a = b.

    Here's my reasoning why 'a <' is defined for all of K.
    First off a < a is defined as false.
    Whenever a < b is defined, a < Sb is defined as a <= b
    Whenever a < Sb is defined, a < b is defined as a < Sb and a = b.

    Assuming that S is a function where this is valid, the previous two
    lines define the predicate P(x) = "a < x" for all x in K.  But there's
    no reason that P(x) defined this way has "not P(a)".  You've added a <
    a as an extra condition; there's no reason that this needs to be
    consistent with the rest of your definition of <.

    Ok, let's remove not a < a.  From a = a, there's a
    base case a < Sa.  For finite K, one has < = KxK.

    However, KxK itself, always satisfies
    a <= b iff a < Sb,
    so the modified definition is useless.

    Definition 2:  <= subset KxK;  a < b when a <= b & a /= b;
    for all a, a <= a;  for all a,b, (a <= b iff a < Sb).

    This has the same problem, namely the
    'definition' is inconsistent for finite K.

    Definition 3:  <= subset KxK;  a < b when a <= b & a /= b;
    for all a, a < Sa;  for all a,b, (a <= b iff a < Sb).

    Again the same problem.  Is there any way to define < or <=
    that's useful and consistent for both finite and infinite sets?


    ...
    However with the establishment of that definition the previous
    finite models create contradictions.  Nothing so simple as < is
    empty or < equals KxK.  No, a raw a < a.

    For example, K = {0,1,2}, S0 = 1;  S1 = 2;  S2 = 0.
    0 <= 0;  0 < S0 = 1;  0 <= 1
    0 < S1 = 2;  0 <= 2;  0 < S2 = 0

    What's wrong with the definition?  
    Where is the blunder in the above reasoning?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)