The Paradox of Symmetric Induction

[William Elliot]

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?

[David Hobby]

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 <.
> ...

[William Elliot]

In article <070320171202351509%ed...@math.ohio-state.edu.invalid>,

David Hobby <david....@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?