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Topic: The Paradox of Symmetric Induction
Replies: 2   Last Post: Mar 12, 2017 3:27 PM

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William Elliot

Posts: 2,637
Registered: 1/8/12
The Paradox of Symmetric Induction
Posted: Mar 6, 2017 11:36 AM
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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

(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?

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