axiom of cardinality in boolean algebra?
Tegiri Nenashi
Propositional calculus. However, I'm not certain if I ever came across
an axiom that explicitly states that there are only 2 values T and F
such as
T | F.
(whatever T and F BA expressions might be). In other words, am I right
to assume that for BA to be faithful representation of Propositional
calculus its model has to have 2 elements?
MoeBlee
Boolean algebra is a first order theory. Boolean algebra has many
models, of many cardinalities. Each model of Boolean algebra is a
Boolean algebra (just as each model of group theory is a group). When
we define a particular Boolean algebra, part of that is stipulating
the carrier set of the algebra, which determines the cardinality of
that carrier set.
For a consistent theory G, the Lindenbaum-Tarski-algebra(G) is a
Boolean algebra, and where G is the set of validities, this is the
Boolean algebra ordinarily taken as the analog of classical
propositional logic.
Going back, start with equivalence classes of formulas modulo the
relation R(G) defined by:
R(G)pq <-> (G is a theory & G is consistent & "p<->q" in G).
Then define the zero element as the equivalence class of logically
false formulas, and the one element as the equivalence class of
logically true formulas. Then define the meet and join operations
accordingly.
If G is a consistent theory, then the result of the above is a Boolean
algebra that is the Tarski-Lindenbaum-algebra(G).
Then take G to be the set of validities. The carrier set of this
Boolean algebra is infinite. Granted, the one and zero elements of
this carrier set are what we take as 'truth' and 'falsehood', but they
are not the only elements, since for each formula p there is the
equivalence class of formulas logically equivalent to p.
MoeBlee
MoeBlee
P.S. In this context, by 'consistent theory' I mean a consistent
theory in a pure propositional language with denumerably many
sentential letters.
MoeBlee
george
> Boolean algebra is a first order theory.
He surely didn't know this.
He thought it was for and about propositional logic.
He may therefore have been thinking in terms of a propositional axiom-
set
for it.
> Boolean algebra has many models, of many cardinalities.
Well, yes but NOT in the sense that HE meant "cardinality".
He wanted to know how you prove that there must be exactly two truth-
values. THAT cardinality (TWO) was the ONLY one HE was talking about.
> Each model of Boolean algebra is a Boolean algebra
> (just as each model of group theory is a group).
Right, but if that's your paradigm, then you need TO STATE SOME
AXIOMS.
There are TWO DIFFERENT relevant sets of axioms here.
One, as you are saying, is 1st-order. The other is 0th-order.
The axiom HE was about was sort of a definition of ~ , something like
~(P<-->~P).
> When we define a particular Boolean algebra, part of that is stipulating
> the carrier set of the algebra, which determines the cardinality of
> that carrier set.
This is an intriguing use of the word "particular".
A "particular" boolean algebra DOESN'T get defined.
The defined concept here IS "boolean algebra" and IT is defined
BY ITS AXIOMS, which DON'T depend on any particular model,
but precisely as you said, define ALL of their models as boolean
algebraS (plural).
The IMPORTANT model (whether at 0th-order or 1st) is the one where
every proposition is a subset of some given set, which can be
identified
with the universe and therefore with any tautological (at 0th order)
proposition
or valid (at 1st-order) sentence.
(at 1st-order) proposition.
Tegiri Nenashi
In order to double check that I'm following let me introduce a sloppy
example. Consider all propositional formulas over two variables X and
Y. Then, there are 16 equivalence classes and I went through a cute
exercise locating formulas like X ^ Y or X ^ -Y v -X ^ Y on 4
dimensional boolean cube.
I guess the question why there are formulas not evaluating to T or F
is outside of propositional calculus/boolean algebra. Is a possible
answer that in predicate calculus we have relations and any non-
zilliary relation can't evaluate to T or F. What about zilliary
relations, however? Can we postulate that zilliary relations always
evaluate to T or F?
MoeBlee
P.P.S.
If my quick calculations are correct (done over a sandwich at lunch),
No matter how many sentential letters in the language of G, the
Lindenbaum-Tarski-algebra(G) has exactly two members in the carrier
set iff G is complete.
And if G is not complete, then, even if the language of G has as
little as one sentential letter, the Lindenbaum-Tarski-algebra(G) has
at least four members in the carrier set.
MoeBlee
MoeBlee
My choice, at this time, not to respond to george in this thread, or
in any other thread, should not be taken as any indication that I
agree or disagree with any particular remarks he posts.
MoeBlee
MoeBlee
> Consider all propositional formulas over two variables X and
> Y. Then, there are 16 equivalence classes
That depends on what G is. If G is the set of validities, then I think
you're right. But I'd want to think it through for myself.
> I guess the question why there are formulas not evaluating to T or F
> is outside of propositional calculus/boolean algebra.
Given any evaluation (assignment of T or of F to each sentential
letter and the evaluation method of truth tables), a formula evaluates
to T or it evaluates to F for that evaluation. But that is a different
matter from the fact that there may be more than two equivalence
classes.
> Is a possible
> answer that in predicate calculus we have relations and any non-
> zilliary relation can't evaluate to T or F. What about zilliary
> relations, however? Can we postulate that zilliary relations always
> evaluate to T or F?
'zilliary' means '0-ary', I surmise.
In predicate logic, aside from 0-ary predicate letters, i.e.,
sentential letters (if your predicate language even includes
sentential letters), n-ary predicate symbols map to n-ary relations on
the universe of the model. Then an atomic formula with an n-ary
predicate symbol evaluates to true iff the tuple named by the string
of terms following the predicate symbol is a member of the relation to
which the predicate symbol maps.
MoeBlee
Tegiri Nenashi
> In predicate logic, aside from 0-ary predicate letters, i.e.,
> sentential letters (if your predicate language even includes
> sentential letters),
There are always sentential formulas (e.g. Ax:P(x)), so the matter if
my predicate language allows sentential letters is not important?
> n-ary predicate symbols map to n-ary relations on
> the universe of the model. Then an atomic formula with an n-ary
> predicate symbol evaluates to true iff the tuple named by the string
> of terms following the predicate symbol is a member of the relation to
> which the predicate symbol maps.
Correct. My question arose in the context of algebraic logic. (For the
record, it is neither Kleene, nor cylindric algebra). I have axiomatic
algebraic system corresponding to predicate calculus, so my question
is how do I interpret the models. The elements sure have to be
relations, and I identified some 0-ary ones. As expected all 0-ary
relatons are be proven top satisfy boolean algebra axioms. I was
wondering if I have to postulate that there are only two such
elements.
Some further investigation (with Mace4), however, revealed that the
arity of the predicate can't be established algebraically. What I
initially thought to be zilliary relations, can as well be interpreted
as unary relations. Then there can be more than two values in the
lattice between lattice top and bottom element...
MoeBlee
> On Feb 3, 3:31 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > In predicate logic, aside from 0-ary predicate letters, i.e.,
> > sentential letters (if your predicate language even includes
> > sentential letters),
>
> There are always sentential formulas (e.g. Ax:P(x)), so the matter if
> my predicate language allows sentential letters is not important?
Some people call those 'prime formulas'. As to importance, yes, my
remark was actually in agreement with what you're saying.
> My question arose in the context of algebraic logic. (For the
> record, it is neither Kleene, nor cylindric algebra). I have axiomatic
> algebraic system corresponding to predicate calculus, so my question
> is how do I interpret the models.
I don't know what you mean by interpret the models. Ordinarily, what
we interpret are languages. By the way, I know about (not details
though) cylindric algebra vis-a-vis predicate logic, but where do I
find out about Kleene's algebra for this?
> The elements sure have to be
> relations, and I identified some 0-ary ones. As expected all 0-ary
> relatons are be proven top satisfy boolean algebra axioms. I was
> wondering if I have to postulate that there are only two such
> elements.
Hmm, I guess I'm starting to see what you're getting at, but I'm not
really sure. But, in a 2-valued logic, what 0-ary relations are there
other than the two?
But let me see whether we're on the same page. Is the following
correct?:
You have a certain 1st order axiomatization. And the models of the
theory thus axiomatized are a certain kind of algebra that in some
sense corresponds to first order logic itself (in some way similar to
Lindenbaum-Tarski algebras corresponding to sentential logic and
cylindric algebras corresponding to1st order logic).
> Some further investigation (with Mace4), however, revealed that the
> arity of the predicate can't be established algebraically. What I
> initially thought to be zilliary relations, can as well be interpreted
> as unary relations. Then there can be more than two values in the
> lattice between lattice top and bottom element...
I don't know what you mean by establishing a predicate algebraically.
Perhaps this is material that I haven't studied yet. I have Halmos's
book on algebraic logic. Do know if this matter is addressed in that
book?
MoeBlee
Tegiri Nenashi
> On Feb 3, 4:07 pm, Tegiri Nenashi <TegiriNena...@gmail.com> wrote:
> > My question arose in the context of algebraic logic. (For the
> > record, it is neither Kleene, nor cylindric algebra). I have axiomatic
> > algebraic system corresponding to predicate calculus, so my question
> > is how do I interpret the models.
>
> I don't know what you mean by interpret the models. Ordinarily, what
> we interpret are languages. By the way, I know about (not details
> though) cylindric algebra vis-a-vis predicate logic, but where do I
> find out about Kleene's algebra for this?
I meant Binary Relation Algebras (BRA) which are often called Relation
Algebras. Kleene's algebra are more generic, so it was sloppy
language on my part.
As far as model interpretation goes consider mace4 system. You feed an
axiom system into it and it outputs a model. This model includes a set
of abstract elements for convenience numbered by nonnegative integers,
and tables of operations ("multiplication" tables). Elements don't
have any structure, so the only fact known about them is what is the
result of operations are applied to them.
However, consider arithmetic as a model for algebra. The elements -
numbers can be interpreted as certain structures (e.g. decimal
representation). Another example is boolean vectors as elements of a
model satisfying Boolean algebra axioms.
In my case, elements are relations of mixed arity (in named, not
positioned, perspective). Therefore, algebraic operations can be
specified in terms of elements structure: for example given a certain
binary operation upon two relations x and y, returning relation z,
what are the attributes of relation z expressed in terms of x and y
attributes, Same question for tuples.
> > The elements sure have to be
> > relations, and I identified some 0-ary ones. As expected all 0-ary
> > relatons are be proven top satisfy boolean algebra axioms. I was
> > wondering if I have to postulate that there are only two such
> > elements.
>
> Hmm, I guess I'm starting to see what you're getting at, but I'm not
> really sure. But, in a 2-valued logic, what 0-ary relations are there
> other than the two?
Variables in the propositional logic language are binary.
> But let me see whether we're on the same page. Is the following
> correct?:
>
> You have a certain 1st order axiomatization. And the models of the
> theory thus axiomatized are a certain kind of algebra that in some
> sense corresponds to first order logic itself (in some way similar to
> Lindenbaum-Tarski algebras corresponding to sentential logic and
> cylindric algebras corresponding to1st order logic).
This is correct. The algebra is lattice which is weaker than boolean
algebra. This is little surprising, because BRA (as a algebraization
of certain fragment of first order logic) is stronger than boolean
algebra.
> > Some further investigation (with Mace4), however, revealed that the
> > arity of the predicate can't be established algebraically. What I
> > initially thought to be zilliary relations, can as well be interpreted
> > as unary relations. Then there can be more than two values in the
> > lattice between lattice top and bottom element...
>
> I don't know what you mean by establishing a predicate algebraically.
> Perhaps this is material that I haven't studied yet. I have Halmos's
> book on algebraic logic. Do know if this matter is addressed in that
> book?
Let's be more specific and write down the system I'm talking about:
x ^ y = y ^ x.
(x ^ y) ^ z = x ^ (y ^ z).
x ^ (x v y) = x.
x v y = y v x.
(x v y) v z = x v (y v z).
x v (x ^ y) = x.
x = (x ^ R00) v (x ^ R11).
(x v (y ^ R00)) ^ (y v (x ^ R00))=(x ^ y) v ((x v y) ^ R00).
x ^ (y' ^ z')' = ((x ^ y)' ^ (x ^ z)')'.
R00 ^ (x ^ (y v z)) = R00 ^ ((x ^ y) v (x ^ z)).
x' ^ x = x ^ R00.
x' v x = x v R11.
R00 and R11 are constants, the "^" and "v" are binary operations, the
quotation mark "'" is an unary one. The variables and constants are
interpreted as relations, so my question is if relation arity can be
deduced from the above axioms. For example, I assumed that in any
interpretation R00 is empty 0-ary relation, but as I wrote in the
previous email it is wrong.
Tegiri Nenashi
> The axiom HE was about was sort of a definition of ~ , something like
> ~(P<-->~P).
Actually, this can be written explicitly to eliminate any confusion
(if any)
x ^ 1 = 1. # definition of lattice top element
x v 0 = 0. # bottom
x = 1 | x = 0. # 2-element Boolean algebra definition
My question was if this axiom is a part of "algebraization" of
propositional calculus.
george
> x ^ 1 = 1. # definition of lattice top element
> x v 0 = 0. # bottom
> x = 1 | x = 0. # 2-element Boolean algebra definition
One way of defining a boolean algebra is as a distributive
complemented lattice.
So (of course) you would need all the axioms for defining a lattice.
Then you could add two, one for complement and one for distributivity.
> My question was if this axiom is a part of "algebraization" of
> propositional calculus.
You need to define what you mean by "algebraization".
It doesn't really matter whether anything is "algebraized" or not.
Boolean algebra is just a theory with some axioms.
All models are legitimate.
Frederick Williams
No. Let the elements of a BA be strings of n 0s and 1s. Operate on
them bit-wise as exemplified by
010001
join 011100
= 011101
to see that there are BAs with 2^n (n a positive integer) elements that
model PC.
--
But you see, I can believe a thing without understanding it.
It's all a matter of training.
--Lord Peter Wimsey in Dorothy L Sayers' _Have His Carcase_
MoeBlee
> I meant Binary Relation Algebras (BRA) which are often called Relation
> Algebras.
Thanks for the rest of your post on that subject. I'm going to read it
again more carefully tonight. I'm only hit and miss in my knowledge of
this particular branch of the subject.
If you're interested though, I'd like to return to one point that came
up regarding Lindenbaum-Tarski.
G is complete iff the cardinality of the carrier set is 2.
And if G is not complete, then with a language of only 1 sentence
letter, the cardinality of the carrier set is 4.
Then you found that that with G not complete and a language of exactly
2 sentence letters, the cardinality of the carrier set is 16.
So my guess is that if G is not complete and the language has n number
of sentence letters (for n a natural number greater than 0), then the
carrier set has cardinaltiy 2^(2^n), which is the number of n-ary
Boolean functions.
So do you know a simple (not involving too much advanced machinery)
proof of that? Is there a simple way to prove your conclustion about
2 sentence letters that generalizes to n number of sentence letters?
Hmm, just need to show that with n sentence letters, the cardinality
of the carrier set is 1-1 with the set of n-ary Boolean functions. Any
ideas (simple ideas, not depending on advanced theorems)?
Thanks,
MoeBlee
Tegiri Nenashi
> If you're interested though, I'd like to return to one point that came
> up regarding Lindenbaum-Tarski.
> G is complete iff the cardinality of the carrier set is 2.
Are you implying that if a (consistent) theory G is complete, then the
model of Lindenbaum-Tarski algebra has domain of cardinality 2?
> And if G is not complete, then with a language of only 1 sentence
> letter, the cardinality of the carrier set is 4.
I'm uncomfortable with theory having syntactic constraints such as
number of allowed variable symbols.
By "not comple" you mean that if language G has variable x, then the
proposition "x" trivially can't be proved nor refuted?
> Then you found that that with G not complete and a language of exactly
> 2 sentence letters, the cardinality of the carrier set is 16.
My very limited familiarity with Lindenbaum-Tarski algebras (via
wikipedia) doesn't allow appreciating this concept. This looks like a
very artificial way of constructing a model for a theory G. What are
main results discovered with its introduction?
> So my guess is that if G is not complete and the language has n number
> of sentence letters (for n a natural number greater than 0), then the
> carrier set has cardinaltiy 2^(2^n), which is the number of n-ary
> Boolean functions.
>
> So do you know a simple (not involving too much advanced machinery)
> proof of that? Is there a simple way to prove your conclustion about
> 2 sentence letters that generalizes to n number of sentence letters?
>
> Hmm, just need to show that with n sentence letters, the cardinality
> of the carrier set is 1-1 with the set of n-ary Boolean functions. Any
> ideas (simple ideas, not depending on advanced theorems)?
Although we speak past each other (like typical cranks do), there was
some utility of this exchange. Lindenbaum-Tarski points to Heyting
algebra page, and it was entertaining to check if Prover9 can deduce
distributivity from Heyting algebra foundation axioms...
MoeBlee
> On Feb 4, 11:23 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > If you're interested though, I'd like to return to one point that came
> > up regarding Lindenbaum-Tarski.
> > G is complete iff the cardinality of the carrier set is 2.
>
> Are you implying that if a (consistent) theory G is complete, then the
> model of Lindenbaum-Tarski algebra has domain of cardinality 2?
Yes; unless I erred in my reckoning, the proof is trivial (though I
don't call the Lindenbaum-Tarski algebra a model). (And I should have
said 'consistent' also in my original formula, but I'm leaving as
tacit in this context that G is consistent since otherwise, it doesn't
induce a Lindenbaum-Tarski algebra.)
> > And if G is not complete, then with a language of only 1 sentence
> > letter, the cardinality of the carrier set is 4.
>
> I'm uncomfortable with theory having syntactic constraints such as
> number of allowed variable symbols.
Whether called 'propositional variables', 'sentential letters', '0-
place predicate symbols', or whatever we wish to call them, it is not
unintelligible to define a PARTICULAR language that has only finitely
many of them. On the other hand, for a first order predicate language,
ordinarily we do equip with denumerably many individual variables
(whether these are each a single symbol or made by adding, say, stroke
marks such as in some formulations of PRA).
> By "not comple" you mean that if language G has variable x, then the
> proposition "x" trivially can't be proved nor refuted?
G is complete in L <-> (G is a set of formulas in the language L &
every formula in the language L is such that either it or its negation
is derivable from G).
Of course, where the language L is understood by context, we may omit
mentioning 'in L' etc.
So, yes, if G is incomplete, then there is a formula that is neither
provable nor refutable from G.
> > Then you found that that with G not complete and a language of exactly
> > 2 sentence letters, the cardinality of the carrier set is 16.
>
> My very limited familiarity with Lindenbaum-Tarski algebras (via
> wikipedia) doesn't allow appreciating this concept. This looks like a
> very artificial way of constructing a model for a theory G. What are
> main results discovered with its introduction?
I'm not constructing a model. Rather, any consistent theory G in a
language of propositional logic induces a certain Lindenbaum-Tarski
algebra. The carrier set of that algebra has a cardinality. My
conjecture is just that if G is incomplete, then the cardinality is 2^
(2^n) where n is the number of sentential letters in the language and
n is finite. Of course, if the number of sentential letters is
denumerable then the cardinality of the carrier set is denumerable
too.
I confirmed this for myself for n=1, and you confirmed it for yourself
(though I don't know how you did it) for n=2.
As to main results, I don't know. I just see how Lindenbaum-Tarski
algebras are an "alter-ego" of propositional logic.
On the other hand, I take an ordinary method of regarding a model for
a theory in a propositional language to be a function from the set of
sentence letters into {0 1} or into {true false} whatever. (Or Chang &
Keisler, even more elegantly, take a model just to be a subset of the
set of sentence letters, as then, intuitively we may regard as true
all the sentence letters that are in the set and false all those not
in the set.)
But given a model, then, of course, there is a corresponding
consistent complete theory, which has as its non-logical axioms just
the sentence letters that are assigned true by the model and the
negations of the sentence letters that are assigned false in the
model. Then the carrier set of the Lindenbaum-Tarski algebra
engendered by that theory has exactly 2 members - the equivalence
class of theorems and the equivalence class of non-theorems.
MoeBlee
William Elliot
One reason that I don't take George, or rather george as he's want to
refer to himself, seriously is that he SHOUTS so much that quick reading
of his posts are unpleasant. As I've already more than I can do in one
evening, those who are nuisances are quickly dismissed unto del-oblivion.
Alan Smaill
> I'm not constructing a model. Rather, any consistent theory G in a
> language of propositional logic induces a certain Lindenbaum-Tarski
> algebra. The carrier set of that algebra has a cardinality. My
> conjecture is just that if G is incomplete, then the cardinality is 2^
> (2^n) where n is the number of sentential letters in the language and
> n is finite. Of course, if the number of sentential letters is
> denumerable then the cardinality of the carrier set is denumerable
> too.
What about the theory with 2 sentential letters A, B
and the single axiom A <-> B ?
> I confirmed this for myself for n=1, and you confirmed it for yourself
> (though I don't know how you did it) for n=2.
The free boolean algebra on n generators has cardinality 2^{2^n},
on the other hand.
http://en.wikipedia.org/wiki/Free_Boolean_algebra
> MoeBlee
>
>
--
Alan Smaill
george
> My question arose in the context of algebraic logic.
In what sense can there be any other kind?
And are you talking about propositional logic or first-order logic?
> (For the record, it is neither Kleene, nor cylindric algebra).
I thought it was BOOLEAN Algebra!?!
>I have axiomatic algebraic system corresponding to predicate calculus,
Are you SURE this is what you want??
If you are talking about boolean algebra (that was what the title of
the
thread said) THEN PROPOSITIONAL and NOT "predicate" calculus was
surely what was relevant! You have to crawl before you walk!
> so my question is how do I interpret the models.
English must not be your native language.
YOU DON'T interpret the models. YOU NEVER interpret the models.
The models are THE OBJECTS, not the subjects, of the interpretations.
You interpret THE FIRST-ORDER LANGUAGE *into or as* a structure,
yielding (if you got it right) a model.
> The elements sure have to be relations,
Not exactly. The "elements" or "objects" ARE THE THINGS IN THE DOMAIN
of the model, the things in THE DOMAIN of the relations "in" the
model.
> and I identified some 0-ary ones.
The 0-ary relations in the model correspond to 0-ary functors in the
language.
This is NOT complicated.
> As expected all 0-ary relatons are be proven top satisfy boolean algebra axioms.
No individual relation satisfies anything.
You are WAY over-complicating this.
You need to go back to the beginning.
A boolean algebra is a lattice but that is almost irrelevant.
What IS relevant is what you need TO ADD to that.
And why do you CARE, ANYway, if you are about FIRST-order
logic or PREDICATE calculus?? That is MUCH harder!
george
> Some further investigation (with Mace4), however, revealed that the
> arity of the predicate can't be established algebraically.
You have not supplied NEARLY enough context here.
The arity OF WHAT predicate?? In almost ANY context in which
ANYONE COULD be doing predicate calculus, WE STARTED
WITH SOME AXIOMS in which the -arity OF EVERY predicate
OR FUNCTOR that the axioms used or mentioned WAS BLATANTLY
OBVIOUS TO THE *EYE*, by inspection, ad oculos. They all have
whatever -arity they are USED with!!
MoeBlee
> MoeBlee <jazzm...@hotmail.com> writes:
> > I'm not constructing a model. Rather, any consistent theory G in a
> > language of propositional logic induces a certain Lindenbaum-Tarski
> > algebra. The carrier set of that algebra has a cardinality. My
> > conjecture is just that if G is incomplete, then the cardinality is 2^
> > (2^n) where n is the number of sentential letters in the language and
> > n is finite. Of course, if the number of sentential letters is
> > denumerable then the cardinality of the carrier set is denumerable
> > too.
>
> What about the theory with 2 sentential letters A, B
> and the single axiom A <-> B ?
Hmm, if I skimmed the combinations correctly, that gives us a carrier
set with cardinality of 4, thus refuting (starting as early as n=2) my
conjecture. Right?
> The free boolean algebra on n generators has cardinality 2^{2^n},
> on the other hand.
Hmm, related to that, I have an idea here I want to ask you about, but
I want to think it through some more first.
MoeBlee
Tegiri Nenashi
> On Feb 3, 7:07 pm, Tegiri Nenashi <TegiriNena...@gmail.com> wrote:
>
> > My question arose in the context of algebraic logic.
>
> In what sense can there be any other kind?
> And are you talking about propositional logic or first-order logic?
My original question arouse in context of algebraization of first-
order logic. I scoped my original question to propositional logic
because everybody agrees what algebraization of propositional calculus
is (boolean algebra). I thought it would be awkward to ask the
question in terms my own algebraization of first-order logic (which
very few people are familiar with).
> > (For the record, it is neither Kleene, nor cylindric algebra).
>
> I thought it was BOOLEAN Algebra!?!
Yes and no, see above.
> >I have axiomatic algebraic system corresponding to predicate calculus,
>
> Are you SURE this is what you want??
> If you are talking about boolean algebra (that was what the title of
> the
> thread said) THEN PROPOSITIONAL and NOT "predicate" calculus was
> surely what was relevant! You have to crawl before you walk!
The thread took a detour from boolean algebra to algebraization of
first-order logic.
> > so my question is how do I interpret the models.
>
> English must not be your native language.
> YOU DON'T interpret the models. YOU NEVER interpret the models.
> The models are THE OBJECTS, not the subjects, of the interpretations.
> You interpret THE FIRST-ORDER LANGUAGE *into or as* a structure,
> yielding (if you got it right) a model.
Would the term "map" make you less upset? As I described in another
message a model is just a structure with elements that are considered
atomic. All what is required for elements is to "fit" into
"multiplication" tables. Yet, one can legitimately wonder if those
elements have structure on their own, so that operations upon them can
be defined in terms of this structure.
> > The elements sure have to be relations,
>
> Not exactly. The "elements" or "objects" ARE THE THINGS IN THE DOMAIN
> of the model, the things in THE DOMAIN of the relations "in" the
> model.
Here is the source of confusion. We have relations/operations in terms
of universal algebra/model theory. We also have relations/predicates
in the scope of predicate calculus. Those have _nothing to do_ with
"relations" in the sentence you quoted. What I'm saying is that
elements or objects in the domain of the model are structured as
relations. Remember, we are not talking about arbitrary model. We
investigating models that correspond to predicate calculus (assuming,
of course, that somebody exhibited universal algebra for predicate
calculus).
> No individual relation satisfies anything.
> You are WAY over-complicating this.
> You need to go back to the beginning.
>
> A boolean algebra is a lattice but that is almost irrelevant.
> What IS relevant is what you need TO ADD to that.
> And why do you CARE, ANYway, if you are about FIRST-order
> logic or PREDICATE calculus?? That is MUCH harder!
I admit I have chosen a wrong venue to promote my algebraization.
Nobody in the right mind would expect writing few lines in a newsgroup
and expect it to have any impact. Therefore, I "cease and desist".
Alan Smaill
> On Feb 5, 8:34 am, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > MoeBlee <jazzm...@hotmail.com> writes:
> > > I'm not constructing a model. Rather, any consistent theory G in a
> > > language of propositional logic induces a certain Lindenbaum-Tarski
> > > algebra. The carrier set of that algebra has a cardinality. My
> > > conjecture is just that if G is incomplete, then the cardinality is 2^
> > > (2^n) where n is the number of sentential letters in the language and
> > > n is finite. Of course, if the number of sentential letters is
> > > denumerable then the cardinality of the carrier set is denumerable
> > > too.
> >
> > What about the theory with 2 sentential letters A, B
> > and the single axiom A <-> B ?
>
> Hmm, if I skimmed the combinations correctly, that gives us a carrier
> set with cardinality of 4, thus refuting (starting as early as n=2) my
> conjecture. Right?
Agreed.
> > The free boolean algebra on n generators has cardinality 2^{2^n},
> > on the other hand.
>
> Hmm, related to that, I have an idea here I want to ask you about, but
> I want to think it through some more first.
>
> MoeBlee
>
--
Alan Smaill