In Barwise's Handbook of Mathematical Logic, CH: A.1, $B!x(B2, it reads
(with some typing ammends):

2.4. Compactness Theorem (Godel-Malcev). Let T be any set of first-
order axioms. If for every finite subset To of T there is a model of
all the axioms in To, then there is a single model of all the axioms
in T.

An alternate form of the Compactness Theorem is sometimes more
convenient. Let us write T |= $B'f(B to indicate that $B'f(B
is a logical consequence of T in the sense that $B'f(B is true in
all models which make all the axioms of T true. Then the Compactness
Theorem is equivalent to the statement: If T U {$B'f(B} is a set
of first order sentences and $B'4(B |= $B'f(B, then there is a
finite subset To of T such that To |= $B'f(B. To see that this
follows from 2.4, apply to 2.4 to T U {not $B'f(B} (sic.), where
"not $B'f(B" asserts that $B'f(B is false. To prove 2.4 from
this version, let $B'f(B be some absurd $B'f(B like 3x(x^x).

I have tried to follow the hints given to prove this equivalence, but
without success.

Hello,





malc...@gmail.com wrote:
In Barwise's Handbook of Mathematical Logic, CH: A.1, $B!x(B2, it reads
(with some typing ammends):

2.4. Compactness Theorem (Godel-Malcev). Let T be any set of first-
order axioms. If for every finite subset To of T there is a model of
all the axioms in To, then there is a single model of all the axioms
in T.

An alternate form of the Compactness Theorem is sometimes more
convenient. Let us write T |= $B'f(B to indicate that $B'f(B
is a logical consequence of T in the sense that $B'f(B is true in
all models which make all the axioms of T true. Then the Compactness
Theorem is equivalent to the statement: If T U {$B'f(B} is a set
of first order sentences and $B'4(B |= $B'f(B, then there is a
finite subset To of T such that To |= $B'f(B. To see that this
follows from 2.4, apply to 2.4 to T U {not $B'f(B} (sic.), where
"not $B'f(B" asserts that $B'f(B is false. To prove 2.4 from
this version, let $B'f(B be some absurd $B'f(B like 3x(x^x).

I have tried to follow the hints given to prove this equivalence, but
without success.

You can use the fact that

T |= phi if and only if T u {not phi} has no model

BTW: Barwise is rather old fashioned (read: outdated). For a good
introduction to logic I would recommend

Ebbinghaus, Flum, Thomas, Mathematical Logic, Springer.

If you are looking for something more advanced, you could try

Marker, Model Theory: An Introduction, Springer.
Hodges, Model Theory, Cambridge University Press.

Achim
--
________________________________________________________________________
| \_____/ |
Achim Blumensath \O/ \___/\ |
TU Darmstadt =o= \ /\ \|
www.mathematik.tu-darmstadt.de/~blumensath /"\ o----|
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On Tue, 22 Apr 2008 03:14:15 -0700 (PDT), malc...@gmail.com wrote:
On 22 abr, 12:00, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
Hello,

malc...@gmail.com wrote:
In Barwise's Handbook of Mathematical Logic, CH: A.1, ?2, it reads
(with some typing ammends):

2.4. Compactness Theorem (Godel-Malcev). Let T be any set of first-
order axioms. If for every finite subset To of T there is a model of
all the axioms in To, then there is a single model of all the axioms
in T.

An alternate form of the Compactness Theorem is sometimes more
convenient. Let us write T |= ? to indicate that ?
is a logical consequence of T in the sense that ? is true in
all models which make all the axioms of T true. Then the Compactness
Theorem is equivalent to the statement: If T U {?} is a set
of first order sentences and ? |= ?, then there is a
finite subset To of T such that To |= ?. To see that this
follows from 2.4, apply to 2.4 to T U {not ?} (sic.), where
"not ?" asserts that ? is false. To prove 2.4 from
this version, let ? be some absurd ? like 3x(x^x).

I have tried to follow the hints given to prove this equivalence, but
without success.

You can use the fact that

  T |= phi  if and only if  T u {not phi} has no model

BTW: Barwise is rather old fashioned (read: outdated). For a good
introduction to logic I would recommend

  Ebbinghaus, Flum, Thomas, Mathematical Logic, Springer.

If you are looking for something more advanced, you could try

  Marker, Model Theory: An Introduction, Springer.
  Hodges, Model Theory, Cambridge University Press.

Achim
--
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I am already using it, however it doesn't take me too far ahead.

Have you tried using the fact that T |= phi  if and only if  
T u {not phi} has no model, as suggested?

David C. Ullrich- Ocultar texto de la cita -

- Mostrar texto de la cita -

On 22 abr, 12:00, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
Hello,





malc...@gmail.com wrote:
In Barwise's Handbook of Mathematical Logic, CH: A.1, ?2, it reads
(with some typing ammends):

2.4. Compactness Theorem (Godel-Malcev). Let T be any set of first-
order axioms. If for every finite subset To of T there is a model of
all the axioms in To, then there is a single model of all the axioms
in T.

An alternate form of the Compactness Theorem is sometimes more
convenient. Let us write T |= ? to indicate that ?
is a logical consequence of T in the sense that ? is true in
all models which make all the axioms of T true. Then the Compactness
Theorem is equivalent to the statement: If T U {?} is a set
of first order sentences and ? |= ?, then there is a
finite subset To of T such that To |= ?. To see that this
follows from 2.4, apply to 2.4 to T U {not ?} (sic.), where
"not ?" asserts that ? is false. To prove 2.4 from
this version, let ? be some absurd ? like 3x(x^x).

I have tried to follow the hints given to prove this equivalence, but
without success.

You can use the fact that

T |= phi if and only if T u {not phi} has no model

BTW: Barwise is rather old fashioned (read: outdated). For a good
introduction to logic I would recommend

Ebbinghaus, Flum, Thomas, Mathematical Logic, Springer.

If you are looking for something more advanced, you could try

Marker, Model Theory: An Introduction, Springer.
Hodges, Model Theory, Cambridge University Press.

Achim
--
________________________________________________________________________
| \_____/ |
Achim Blumensath \O/ \___/\ |
TU Darmstadt =o= \ /\ \|
www.mathematik.tu-darmstadt.de/~blumensath /"\ o----|
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I am already using it, however it doesn't take me too far ahead.

On 22 abr, 12:00, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
malc...@gmail.com wrote:
2.4. Compactness Theorem (Godel-Malcev). Let T be any set of first-
order axioms. If for every finite subset To of T there is a model of
all the axioms in To, then there is a single model of all the axioms
in T.

An alternate form of the Compactness Theorem is sometimes more
convenient. Let us write T |= ? to indicate that ?
is a logical consequence of T in the sense that ? is true in
all models which make all the axioms of T true. Then the Compactness
Theorem is equivalent to the statement: If T U {?} is a set
of first order sentences and ? |= ?, then there is a
finite subset To of T such that To |= ?. To see that this
follows from 2.4, apply to 2.4 to T U {not ?} (sic.), where
"not ?" asserts that ? is false. To prove 2.4 from
this version, let ? be some absurd ? like 3x(x^x).

I have tried to follow the hints given to prove this equivalence, but
without success.

You can use the fact that

  T |= phi  if and only if  T u {not phi} has no model

T |= phi if and only if there is no model of T u {not phi}. (By
definition of T |= phi)

There is no model of T u {not phi} if and only if some finite subset
of T u {not phi} that does have no model. (By 2.4).

But now, what? I should be able to prove that there is some subset To
of T such that every model of To makes phi become true.

Hello,





malc...@gmail.com wrote:
On 22 abr, 12:00, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
malc...@gmail.com wrote:
2.4. Compactness Theorem (Godel-Malcev). Let T be any set of first-
order axioms. If for every finite subset To of T there is a model of
all the axioms in To, then there is a single model of all the axioms
in T.

An alternate form of the Compactness Theorem is sometimes more
convenient. Let us write T |= ? to indicate that ?
is a logical consequence of T in the sense that ? is true in
all models which make all the axioms of T true. Then the Compactness
Theorem is equivalent to the statement: If T U {?} is a set
of first order sentences and ? |= ?, then there is a
finite subset To of T such that To |= ?. To see that this
follows from 2.4, apply to 2.4 to T U {not ?} (sic.), where
"not ?" asserts that ? is false. To prove 2.4 from
this version, let ? be some absurd ? like 3x(x^x).

I have tried to follow the hints given to prove this equivalence, but
without success.

You can use the fact that

  T |= phi  if and only if  T u {not phi} has no model

T |= phi if and only if there is no model of T u {not phi}. (By
definition of T |= phi)

There is no model of T u {not phi} if and only if some finite subset
of T u {not phi} that does have no model. (By 2.4).

But now, what? I should be able to prove that there is some subset To
of T such that every model of To makes phi become true.

You start at the wrong place. You want to prove that

  T |= phi iff T_0 |= phi, for some finite T_0

One direction is obvious (why?). For the other one:

You assume that T |= phi and you want to show that T_0 |= phi, for some
finite T_0. This follows immediately form the above hint.

Achim
--
________________________________________________________________________
                                                              | \_____/ |
   Achim Blumensath                                          \O/ \___/\ |
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On 23 abr, 08:17, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
malc...@gmail.com wrote:
T |= phi if and only if there is no model of T u {not phi}. (By
definition of T |= phi)

There is no model of T u {not phi} if and only if some finite subset
of T u {not phi} that does have no model. (By 2.4).

But now, what? I should be able to prove that there is some subset To
of T such that every model of To makes phi become true.

You start at the wrong place. You want to prove that

  T |= phi iff T_0 |= phi, for some finite T_0

One direction is obvious (why?). For the other one:

You assume that T |= phi and you want to show that T_0 |= phi, for some
finite T_0. This follows immediately form the above hint.

Statement A: T |= phi iff T_0 |= phi, for some finite T_0

Every model of T is a model of T_0, hence the easy direction (if for
some finite T_0 |= phi then T |= phi).

The same easy direction is present at:

Statement B: For every finite subset To of T there is a model of
all the axioms in To iff there is a single model of all the axioms in
T.

I am trying to prove the equivalence of the 'not that easy' direction
of both statements.

Hello,





malc...@gmail.com wrote:
On 23 abr, 08:17, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
malc...@gmail.com wrote:
T |= phi if and only if there is no model of T u {not phi}. (By
definition of T |= phi)

There is no model of T u {not phi} if and only if some finite subset
of T u {not phi} that does have no model. (By 2.4).

But now, what? I should be able to prove that there is some subset To
of T such that every model of To makes phi become true.

You start at the wrong place. You want to prove that

  T |= phi iff T_0 |= phi, for some finite T_0

One direction is obvious (why?). For the other one:

You assume that T |= phi and you want to show that T_0 |= phi, for some
finite T_0. This follows immediately form the above hint.

Statement A: T |= phi iff T_0 |= phi, for some finite T_0

Every model of T is a model of T_0, hence the easy direction (if  for
some finite T_0 |= phi then T |= phi).

The same easy direction is present at:

Statement B: For every finite subset To of T there is a model of
all the axioms in To iff there is a single model of all the axioms in
T.

I am trying to prove the equivalence of the 'not that easy' direction
of both statements.

    T |= phi

=>  T u {not phi} has no model

=>  ex. finite subset S of T u {not phi} that has no model

Let T_0 := S - {not phi}.

Claim: T_0 |= phi

Does this get you started?

Achim
--
________________________________________________________________________
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   Achim Blumensath                                          \O/ \___/\ |
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On 22 abr, 14:29, David C. Ullrich <dullr...@sprynet.com> wrote:
On Tue, 22 Apr 2008 03:14:15 -0700 (PDT), malc...@gmail.com wrote:
On 22 abr, 12:00, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
Hello,

malc...@gmail.com wrote:
In Barwise's Handbook of Mathematical Logic, CH: A.1, ?2, it reads
(with some typing ammends):

2.4. Compactness Theorem (Godel-Malcev). Let T be any set of first-
order axioms. If for every finite subset To of T there is a model of
all the axioms in To, then there is a single model of all the axioms
in T.

An alternate form of the Compactness Theorem is sometimes more
convenient. Let us write T |= ? to indicate that ?
is a logical consequence of T in the sense that ? is true in
all models which make all the axioms of T true. Then the Compactness
Theorem is equivalent to the statement: If T U {?} is a set
of first order sentences and ? |= ?, then there is a
finite subset To of T such that To |= ?. To see that this
follows from 2.4, apply to 2.4 to T U {not ?} (sic.), where
"not ?" asserts that ? is false. To prove 2.4 from
this version, let ? be some absurd ? like 3x(x^x).

I have tried to follow the hints given to prove this equivalence, but
without success.

You can use the fact that

  T |= phi  if and only if  T u {not phi} has no model

BTW: Barwise is rather old fashioned (read: outdated). For a good
introduction to logic I would recommend

  Ebbinghaus, Flum, Thomas, Mathematical Logic, Springer.

If you are looking for something more advanced, you could try

  Marker, Model Theory: An Introduction, Springer.
  Hodges, Model Theory, Cambridge University Press.

Achim
--
________________________________________________________________________
                                                              | \_____/ |
   Achim Blumensath                                          \O/ \___/\ |
   TU Darmstadt                                              =o=  \ /\ \|
   www.mathematik.tu-darmstadt.de/~blumensath              /"\   o----|
____________________________________________________________________\___|- Ocultar texto de la cita -

- Mostrar texto de la cita -

I am already using it, however it doesn't take me too far ahead.

Have you tried using the fact that T |= phi  if and only if  
T u {not phi} has no model, as suggested?

David C. Ullrich- Ocultar texto de la cita -

- Mostrar texto de la cita -

Why don't you believe me? Is it so obvious?

T |= phi if and only if there is no model of T u {not phi}. (By
definition of T |= phi)

There is no model of T u {not phi} if and only if some finite subset
of T u {not phi} that does have no model. (By 2.4).

But now, what? I should be able to prove that there is some subset To
of T such that every model of To makes phi become true.

Hello,





malc...@gmail.com wrote:
In Barwise's Handbook of Mathematical Logic, CH: A.1, $B!x(B2, it reads
(with some typing ammends):

2.4. Compactness Theorem (Godel-Malcev). Let T be any set of first-
order axioms. If for every finite subset To of T there is a model of
all the axioms in To, then there is a single model of all the axioms
in T.

An alternate form of the Compactness Theorem is sometimes more
convenient. Let us write T |= $B'f(B to indicate that $B'f(B
is a logical consequence of T in the sense that $B'f(B is true in
all models which make all the axioms of T true. Then the Compactness
Theorem is equivalent to the statement: If T U {$B'f(B} is a set
of first order sentences and $B'4(B |= $B'f(B, then there is a
finite subset To of T such that To |= $B'f(B. To see that this
follows from 2.4, apply to 2.4 to T U {not $B'f(B} (sic.), where
"not $B'f(B" asserts that $B'f(B is false. To prove 2.4 from
this version, let $B'f(B be some absurd $B'f(B like 3x(x^x).

I have tried to follow the hints given to prove this equivalence, but
without success.

You can use the fact that

T |= phi if and only if T u {not phi} has no model

BTW: Barwise is rather old fashioned (read: outdated). For a good
introduction to logic I would recommend

Ebbinghaus, Flum, Thomas, Mathematical Logic, Springer.

If you are looking for something more advanced, you could try

Marker, Model Theory: An Introduction, Springer.
Hodges, Model Theory, Cambridge University Press.

Achim
--
________________________________________________________________________
| \_____/ |
Achim Blumensath \O/ \___/\ |
TU Darmstadt =o= \ /\ \|
www.mathematik.tu-darmstadt.de/~blumensath /"\ o----|
____________________________________________________________________\___|- Ocultar texto de la cita -

- Mostrar texto de la cita -

On 22 abr, 12:00, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
BTW: Barwise is rather old fashioned (read: outdated). For a good
introduction to logic I would recommend

Ebbinghaus, Flum, Thomas, Mathematical Logic, Springer.

If you are looking for something more advanced, you could try

Marker, Model Theory: An Introduction, Springer.
Hodges, Model Theory, Cambridge University Press.

Why do you believe Barwise's Handbook is outdated?
I am also reading Ebbinghaus et al. but I find Barwise gives a broader
view of the subject, doesn't it?

malc...@gmail.com wrote:
On 22 abr, 12:00, Achim Blumensath <blumens...@mathematik.tu-
darmstadt.de> wrote:
BTW: Barwise is rather old fashioned (read: outdated). For a good
introduction to logic I would recommend

  Ebbinghaus, Flum, Thomas, Mathematical Logic, Springer.

If you are looking for something more advanced, you could try

  Marker, Model Theory: An Introduction, Springer.
  Hodges, Model Theory, Cambridge University Press.

Why do you believe Barwise's Handbook is outdated?
I am also reading Ebbinghaus et al. but I find Barwise gives a broader
view of the subject, doesn't it?

Yes, Barwise covers more material. But the book was written in the 70s
and much has changed since then. I'm not that familiar with proof
theory, but for model theory I really cannot recommend the book of
Barwise.

Achim
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