Re: Existence, Self-identity and Uniqueness.

On Thu, 28 Dec 2006, Jan Burse wrote:
William Elliot wrote:

Ie, when E!x F(x)
So I see not the objection.

The objection is that "x=y -> (A(x)<->A(y))"
should always hold in FOL=, and not only
under some circumstances.

If it holds only under some circustances, i.e. substituting for x the
expression "the x:Fx", picking the A that I gave (call it A0), and
picking and F that is not function (call it F0), then we arrive at a
contradiction.

The contraction is seen immediately as
follows. In FOL= we have the following
axiom schema:

FOL= |- forall x forall y(x=y -> (A(x)<->A(y))) (1)

The counterexample from my previous email states:

FOL= |- ~ forall y (the x:F0x -> (A0(the x:F0x) <-> A0(y)) (2)

That is not wff because ixF0(x) is a syntatic constant requiring
to be incorporated into a predicate.

Now specialize (1) by the x:F0x and A0, and one gets:
FOL= |- forall y(the x:F0x -> (A0(the x:F0x)<->A0(y)) (3)

The wff with the syntatic iota constant
ixF(x) = y -> (A(ixF(x)) <-> A(y))

is theorem no matter what F and A
where G(ixF(x)) is definition for
(Ex)G(x) & (Ax,y)(F(x) & F(y) -> x = y)

We agree that it's theorem when (E!x)F(x).
In the event ~(E!x)F(x), the premis
ixF(x) = y
is false, thus promoting again the whole statement to truthood.

Thus we have with (2) and (3) that FOL= is inconsistent.
But lets assume that FOL= is not inconsistent, so
the problem is the form "the x:Fx". It renders
FOL= inconsistent.

Usually we want logical forms that do not render our
logic FOL= inconsistent. Because in classical logic,
an inconsistency allows us to derive any formula,
which makes the logic total useless.

Syntatic bothers are deciding if
ixF(x) = ixF(x)
is one application of the definition or not and if
ixF(x) = iyF(y)
is two applications of the definition or not.

One also needs to show
H(G(ixF(x))
is the same as or equivalent to
HG(ixF(x))
where HG is the predicate H(G(x)).

Also that
H & G(ixF(x))
and
K(ixF(x))
are equivalent where K(x) is H & G(x)
and so forth for other variants over the logical connectives.

More bother ensues with quantifies, to show the definition is
invariant over quantifiers or even for H(ixF(x), ixG(x)) that
the order of applying the definition to F and G grants identical
or equivalent results.

.

Relevant Pages

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