...
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.
...
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.
...
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?
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?
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