Re: FO logic without equality



David C. Ullrich wrote:
So in other words, when you say that something
can be found in every book on logic you don't
actually have a book in mind where it can be
found? Similarly when you say that something
is an exercsise in every textbook, you don't
actually know of a single text where it _is_
an exercise?
Wait. For FOL+= ~ FOL+EQ see:

See for example Richter, M.M.: Logikkalküle,
Teubner Studienbücher, Informatik, 1978

He calls a normal model, a "Gleichheitsstruktur".

3.4 Prädikatenlogik mit Gleichheit (FOL with =)
3. Satz: Sei Sigma eine Formelmenge in der Gleichheitslogik,
AA ein gewöhnliches (d.h. nicht notwendig Gleichheits-)
Modell von Sigma [and the = Axioms], R subset A^2 sei
die Interpretation von "=". Dann ist R eine Kongruenzrelation
in AA und AA/R ist ein Gleichheits modell für Sigma.

(For a set Sigma of formulae in FOL with =, if AA is
an ordinary (i.e. not necessarily a normal-) model
of Sigma [and the = Axioms], and let R subset A^2 be the
interpretation of "=". Then R is a congruence relation
in AA and AA/R is a normal model for Sigma.)

> I'd really like to know what textbook
> you find these "exercises" in. What's
> a sentence phi_k in FOL _without_
> equality that says every model has
> cardinality k?
Slight miss understanding, I meant FOL with =.

Ebbinghaus, H.-D., Flum, J. and Thomas, W.: Mathematical
Logic, Second Edition, 1994

III. Semantics of First-Order Languages
6.3 Cardinality Statements. The sentence
phi>=2 := Ex Ey ~x=y
is a formalization of "there are at least two elements.",
[...]. In a similar way, for n>=3, the sentence
phi>=n := Ex1 .. Exn-1(~x0=x1 & ... & ~x0=xn-1 & ... &
~xn-2=xn-1) states that there at least n elements.

Just use the sentence
phi_k := phi>=k & ~phi>=k+1

In FOL without =, phi>=k rewritten to EQ and
the EQ Axioms should also work, but phi_k
is problematic.

Bye
.