Re: For All x

On Mar 27, 11:58 am, apoorv <sudhir...@xxxxxxxxxxx> wrote:
Def: x is finite <-> (x=0 v (~x=0 & Ey(y is finite & x=Sy & ~y=x)))

NO GOOD. CIRCULAR DEFINITION. 'finite' cannot be defined by a right
side of the biconditional that has an occurrence of 'finite'.

'x is finite' is defined in terms of '  predecessor of x ' being
finite; definition is recursive and
not circular.

NO, WRONG. Recursive definitions are justified by "definition by
recursion" theorems. You have no such theorem yet shown in your
theory. Your remark shows a fundamental ignorance and confusion about
recursive definition. This is another good example of your need to
learn the basics of this subject from a textbook.

Now, as to fixing the problem in your system, my guess is that your
system will prove the needed definition by recursion theorem (or even
inductivity, since you don't have unions or intersections, etc. that
are part of the ordinary machinery of induction). So instead you could
take 'finite' as primitive then state your "definition" as an axiom
rather than as a definition. I don't know whether that would empower
your system in the way you want, but at least it would be coherent.

In sum, NO, you cannot define 'finite' by referring to 'finite', and
you are not correct to conflate this situation with one in which we
prove various defintion by recursion theorems then utilize those to
prove the existence of certain "recursively (and/or inductively)
defined" functions and predicates.

However, There is need for an axiom which will prevent an unending
chain of predecessors
Axiom: ~Ez Ay [y e z-->(Ex : ~x=y and Sx =y)]
It can be seen that any x  other than 0 which has no predecessor is
not finite.

I'm going to cut you off there. You used the predicate 'finite'. Get
back to me when you've given a proper definition of 'finite' from your
primitives.

MoeBlee

.

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