Jan Brosius writes:
 > I must put this in the good way;
 > 
 > [forall x . alpha(x)] => alpha(x) is True

Yes, by instantiation.

 > If alpha(x) is TRUE then the following is true : alpha(x) => [forall x.
 > alpha(x)]

No, alpha(x) only asserts that some element named x satisfies alpha. It does
not follow that therefore every other element also satisfies alpha.

 > So if alpha(x) is TRUE then [forall x. alpha(x) ]<=> alpha(x)

This does not follow. It is truth-equivalent to:

  [forall x. alpha(x)] <=> True

which is equivalent to:

  forall x. alpha(x)

which is only true when every element satisifies alpha; so it is not a
tautology.

I repeat: you do not understand the difference between bound and free
variables. A free variable behaves like a constant. It is not "implicitly"
quantified in any way, because it denotes a specific element. The only
difference between a constant and a free variable is syntactic: a free
variable can be bound in an outer scope, while a constant cannot.

 > > The sentence "alpha(x)" asserts that there is a _distinguished_ element
 > > named
 > 
 > NO , saying that there is a distinguished element T that satisfies alpha
 > implies
 > 
 > that exists x. alpha(x) is true

Yes, it also implies this fact, because it can be derived by extensionality:

  alpha(x) => exists y. alpha(y)

However, existential quantification hides the identity of the element in
question. The fact that we know _which_ element it is that satisifies alpha
permits us to say more. This is why:

  exists x. alpha(x),
  alpha(x),

and

  forall x. alpha(x)

are not truth-equivalent. Rather we have:

  forall y. alpha(y) => alpha(x)   and   alpha (x) => exists z. alpha(z)

 > > "x" in the domain of your model, and that that element, at least,
 > satisfies
 > 
 > x is a variable ; no domain ; no model

A model must assign an element in the domain to each free variable. You need a
model to determine the truth-value of a first-order proposition which is not
tautological. We are trying to establish the truth-value of a proposition with
a free variable; therefore we need a model. A model needs a domain of elements
to draw from. Therefore we need a domain. OK?

-- 
Frank Atanassow, Dept. of Computer Science, Utrecht University Padualaan 14,
PO Box 80.089, 3508 TB Utrecht, Netherlands Tel +31 (030) 253-1012, Fax +31
(030) 251-3791


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