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| marcos... |
 Posted: Mon Nov 02, 2009 4:06 am |
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Hello everybody. I am amateur in logic and I am reading "Logic for
mathematicians", written by A.G. Hamilton. I have a doubt:
In the proposition number 4.38, at the end, it says: "...we will
restrict "I" excluding interpretations of the constants b0, b1, ...
and the terms related to them, but we will let "DI" with no
modification. This will bring us an interpretation of "L" in wich
every theorem of "S" is true."
My doubt is this:
"DI" includes all closed terms of "L+", so it can not be an
interpretation of "L".
I must be wrong, but don't know wich is the mistake. |
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| marcos... |
| Posted: Tue Nov 03, 2009 10:40 pm |
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On 4 nov, 07:44, William Elliot <ma... at (no spam) rdrop.remove.com> wrote:
[quote]On Mon, 2 Nov 2009, marcos wrote:
Hello everybody. I am amateur in logic and I am reading "Logic for
mathematicians", written by A.G. Hamilton. I have a doubt:
In the proposition number 4.38, at the end, it says: "...we will
restrict "I" excluding interpretations of the constants b0, b1, ...
and the terms related to them, but we will let "DI" with no
modification. This will bring us an interpretation of "L" in wich
every theorem of "S" is true."
My doubt is this:
"DI" includes all closed terms of "L+", so it can not be an
interpretation of "L".
I must be wrong, but don't know wich is the mistake.
I doubt that you understand everybody isn't like you
with a copy of the text to understand your question.
[/quote]
Sorry, William. I've written the translation I did from spanish, and
without the book is impossible to understand the question. I will
probably buy the book in english, and send to this group more
information about my doubts. |
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| William Elliot... |
| Posted: Wed Nov 04, 2009 1:44 am |
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On Mon, 2 Nov 2009, marcos wrote:
[quote]Hello everybody. I am amateur in logic and I am reading "Logic for
mathematicians", written by A.G. Hamilton. I have a doubt:
In the proposition number 4.38, at the end, it says: "...we will
restrict "I" excluding interpretations of the constants b0, b1, ...
and the terms related to them, but we will let "DI" with no
modification. This will bring us an interpretation of "L" in wich
every theorem of "S" is true."
My doubt is this:
"DI" includes all closed terms of "L+", so it can not be an
interpretation of "L".
I must be wrong, but don't know wich is the mistake.
I doubt that you understand everybody isn't like you[/quote]
with a copy of the text to understand your question. |
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| marcos... |
| Posted: Wed Nov 04, 2009 11:03 pm |
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Guest
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Now I will try to explain the doubt for everybody: those who have read
the book and those who have not.
The proposition the book tries to prove is this:
" If S is a consistent extension of K ( K is a deduction system with a
language L, some axioms, and some deduction rules. This proposition is
the step before proving that K contains all logicaly valid formulas of
L ), then there is an interpretation of L in wich every theorem of S
is true".
To achieve this, the book needs a complete and consistent extension of
closed formulas. Until this moment there were no constants on L. Now
it adds an infinite list of new constants b0, b1, b2, .... The new
language is called L+. L+ is needed to obtain a complete and
consistent extension of S, that will be called T.
Once obtained, it will define an interpretation I of L+ like this:
1- The domain DI will be the set of all closed terms of L+
( individual constants and all the terms built using the constants)
2- The individual constants are their own interpretations
3- Given d1, d2,... that belong to DI, the interpretation of the
functions recieves the value f(d1,...,dn). Funtions will also be
closed terms, of course.
After explaining this interpretation I, the book proves that every
theorem of T is true in that interpretation I.
Now it comes the doubt: Every theorem of S is theorem also of T, so
every formula of L+ that is theorem of S is true in I. Of course,
every theorem of S is a formula of L, and I contains interpretations
of formulas that don't belong to L, so we will restrict I excluding
interpretations of constants b0, b1, ... and the terms related to
them, but we will let DI with no modification. This will bring us an
interpretation of L in wich every theorem of S true.
This is the doubt: DI still includes all closed terms of L+, so ¿isn't
it still an interpretation of L+?.
THANKS EVERYBODI. |
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| marcos... |
| Posted: Wed Nov 04, 2009 11:09 pm |
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On 5 nov, 10:03, marcos <mar... at (no spam) tomasacarolina.e.telefonica.net>
wrote:
[quote]Now I will try to explain the doubt for everybody: those who have read
the book and those who have not.
The proposition the book tries to prove is this:
" If S is a consistent extension of K ( K is a deduction system with a
language L, some axioms, and some deduction rules. This proposition is
the step before proving that K contains all logicaly valid formulas of
L ), then there is an interpretation of L in wich every theorem of S
is true".
To achieve this, the book needs a complete and consistent extension of
closed formulas. Until this moment there were no constants on L. Now
it adds an infinite list of new constants b0, b1, b2, .... The new
language is called L+. L+ is needed to obtain a complete and
consistent extension of S, that will be called T.
Once obtained, it will define an interpretation I of L+ like this:
1- The domain DI will be the set of all closed terms of L+
( individual constants and all the terms built using the constants)
2- The individual constants are their own interpretations
3- Given d1, d2,... that belong to DI, the interpretation of the
functions recieves the value f(d1,...,dn). Funtions will also be
closed terms, of course.
After explaining this interpretation I, the book proves that every
theorem of T is true in that interpretation I.
Now it comes the doubt: Every theorem of S is theorem also of T, so
every formula of L+ that is theorem of S is true in I. Of course,
every theorem of S is a formula of L, and I contains interpretations
of formulas that don't belong to L, so we will restrict I excluding
interpretations of constants b0, b1, ... and the terms related to
them, but we will let DI with no modification. This will bring us an
interpretation of L in wich every theorem of S true.
This is the doubt: DI still includes all closed terms of L+, so ¿isn't
it still an interpretation of L+?.
THANKS EVERYBODI.
[/quote]
SORRY, when i've defined I, I've forgoten one point:
4- Given d1, d2, ... that belong to DI, the formulas are verified if
they are theorems of T |
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| Begoña Castillo... |
| Posted: Thu Nov 05, 2009 11:00 pm |
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Guest
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On 6 nov, 08:05, William Elliot <ma... at (no spam) rdrop.remove.com> wrote:
[quote]On Thu, 5 Nov 2009, marcos wrote:
" If S is a consistent extension of K ( K is a deduction system with a
language L, some axioms, and some deduction rules. This proposition is
the step before proving that K contains all logicaly valid formulas of
L ), then there is an interpretation of L in wich every theorem of S
is true".
To achieve this, the book needs a complete and consistent extension of
closed formulas. Until this moment there were no constants on L. Now
it adds an infinite list of new constants b0, b1, b2, .... The new
language is called L+. L+ is needed to obtain a complete and
consistent extension of S, that will be called T.
Once obtained, it will define an interpretation I of L+ like this:
1- The domain DI will be the set of all closed terms of L+
( individual constants and all the terms built using the constants)
2- The individual constants are their own interpretations
3- Given d1, d2,... that belong to DI, the interpretation of the
functions recieves the value f(d1,...,dn). Funtions will also be
closed terms, of course.
4- Given d1, d2, ... that belong to DI, the formulas are verified if
they are theorems of T
After explaining this interpretation I, the book proves that every
theorem of T is true in that interpretation I.
Now it comes the doubt: Every theorem of S is theorem also of T, so
every formula of L+ that is theorem of S is true in I. Of course,
every theorem of S is a formula of L, and I contains interpretations
of formulas that don't belong to L, so we will restrict I excluding
interpretations of constants b0, b1, ... and the terms related to
them, but we will let DI with no modification. This will bring us an
interpretation of L in which every theorem of S true.
This is the doubt: DI still includes all closed terms of L+, so isn't
it still an interpretation of L+?.
If by it, do you mean the restricted I of the previous paragraphy,
then no, it isn't because the new constants aren't assigned values.
----- Ocultar texto de la cita -
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[/quote]
Thanks, William. I will probably have more questions in the future, so
you will find me again in this group. Thanks. |
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| William Elliot... |
| Posted: Fri Nov 06, 2009 2:05 am |
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Guest
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On Thu, 5 Nov 2009, marcos wrote:
[quote]" If S is a consistent extension of K ( K is a deduction system with a
language L, some axioms, and some deduction rules. This proposition is
the step before proving that K contains all logicaly valid formulas of
L ), then there is an interpretation of L in wich every theorem of S
is true".
To achieve this, the book needs a complete and consistent extension of
closed formulas. Until this moment there were no constants on L. Now
it adds an infinite list of new constants b0, b1, b2, .... The new
language is called L+. L+ is needed to obtain a complete and
consistent extension of S, that will be called T.
Once obtained, it will define an interpretation I of L+ like this:
1- The domain DI will be the set of all closed terms of L+
( individual constants and all the terms built using the constants)
2- The individual constants are their own interpretations
3- Given d1, d2,... that belong to DI, the interpretation of the
functions recieves the value f(d1,...,dn). Funtions will also be
closed terms, of course.
[/quote]
4- Given d1, d2, ... that belong to DI, the formulas are verified if
they are theorems of T
[quote]After explaining this interpretation I, the book proves that every
theorem of T is true in that interpretation I.
Now it comes the doubt: Every theorem of S is theorem also of T, so
every formula of L+ that is theorem of S is true in I. Of course,
every theorem of S is a formula of L, and I contains interpretations
of formulas that don't belong to L, so we will restrict I excluding
interpretations of constants b0, b1, ... and the terms related to
them, but we will let DI with no modification. This will bring us an
interpretation of L in which every theorem of S true.
This is the doubt: DI still includes all closed terms of L+, so isn't
it still an interpretation of L+?.
[/quote]
If by it, do you mean the restricted I of the previous paragraphy,
then no, it isn't because the new constants aren't assigned values.
---- |
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