• #### 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
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• 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)