Stuck on modal logic proof
Rotwang
I'm currently reading "Vicious Circles" by Barwise and Moss, and am
stuck on a proof in Chapter 11, on modal logic. This is the first
thing I've read about modal logic, so I apologise if my question
appears stupid. I'll also give a bit of background since I have no
idea how standard the authors' definitions and notations are: the
system in question has a language L whose sentences include T and a
set of A atomic propositions, and are closed under the connectives ¬,
&, <> and []. The axioms of the system are all substitution instances
of theorems of zeroth order logic (including T) and all substitution
instances of
[](p -> q) -> (([]p) -> ([]q)).
The rules of inference are Modus Ponens,
p p -> q
-------------
q
and necessitation,
p
-------------
[]p.
If p is a theorem we write
|- p.
More generally, let S be a set of sentences. We write
S |- p
iff there is a finite sequence of sentences p_1, p_2, ... p_n in S
such that
|- (p_1 & p_2 & ... p_n) -> p.
Suppose now that U is a maximal consistent set of sentences (where I
think that "consistent" is supposed to mean that there is no sentence
p such that U |- p and U |- ¬p, and maximal means that U u {p} is
inconsistent whenever p is not in U), and that <>p is in U for some p.
Call a set S of sentences "good" if p in S, and whenever R is a finite
subset of S then <> & R is in U (here & R means r_1 & r_2 & ... r_n,
where r_i are the elements of R). The authors have proved that there
is a maximal good set V (maximal wrt being good), and that V is
consistent, and are now trying to prove that V is also maximal wrt
consistency. Suppose otherwise: then there is some q such that neither
q nor ¬q belong to V. So neither V u {q} nor V u {¬q} are good. The
authors then claim that this implies that there is a finite W c V such
that |- & W -> q and |- & W -> ¬q, and use this to derive a
contradiction. But I don't see how this follows. In particular the
fact that V has already been shown to be consistent seems to imply
that no such W exists (and this isn't the contradiction they give -
rather they work a bit harder to give a different contradiction).
Clearly I've misunderstood what's going on - can anybody help?
Having tried to prove that V is maximal wrt consistency myself, the
best I've managed is to show that, with q as above, we have that both
<>q and <>¬q are in U. But I don't think that's enough for a
contradiction.
Rotwang
round. Sorry.
William Elliot
> system in question has a language L whose sentences include T and a
> set of A atomic propositions, and are closed under the connectives 1/4,
What do you mean by 1/4? Negation or ~ as rendered in ascii.
Use plane text as characters that aren't plane text as you see
on the top of the keys, don't show up as you expect to all readers.
> &, <> and []. The axioms of the system are all substitution instances
> of theorems of zeroth order logic (including T) and all substitution
> instances of
>
> [](p -> q) -> (([]p) -> ([]q)).
>
> The rules of inference are Modus Ponens,
>
> p p -> q
> -------------
> q
>
> and necessitation,
>
> p
> -------------
> []p.
>
Are you sure that's a rule of inference for modal logic?
Seems that would imply []p <-> p, that []p is same as p.
> If p is a theorem we write
>
> |- p.
>
> More generally, let S be a set of sentences. We write
>
> S |- p
>
> iff there is a finite sequence of sentences p_1, p_2, ... p_n in S
> such that
>
> |- (p_1 & p_2 & ... p_n) -> p.
>
> Suppose now that U is a maximal consistent set of sentences (where I
> think that "consistent" is supposed to mean that there is no sentence
> p such that U |- p and U |- 1/4 p, and maximal means that U u {p} is
> inconsistent whenever p is not in U), and that <>p is in U for some p.
Maximall consistent means that if p not in U, the U \/ {p} is
inconsistent. In modal logic, I'm not so sure. You seem to be
unsure also. Do you have an exact quote from a text to give?
In your statement p seems to be quantified twice. Once when first
mentioned and again at the end. Thus the meaning is not clear.
> Call a set S of sentences "good" if p in S, and whenever R is a finite
> subset of S then <> & R is in U (here & R means r_1 & r_2 & ... r_n,
> where r_i are the elements of R).
Huh? Is there any connection between U and S? **
> The authors have proved that there is a maximal good set V (maximal wrt
> being good), and that V is consistent, and are now trying to prove that
> V is also maximal wrt consistency. Suppose otherwise: then there is some
> q such that neither q nor 1/4 q belong to V. So neither V u {q} nor V u
> {1/4 q} are good.
> The authors then claim that this implies that there is a finite W c V
What does the term W c V mean? Is c a binary operator? Anything wrong
with being clear and writing W subset V. Is that what you mean?
> such that |- & W -> q and |- & W -> 1/4 q, and use this to derive a
> contradiction. But I don't see how this follows. In particular the fact
> that V has already been shown to be consistent seems to imply that no
> such W exists (and this isn't the contradiction they give - rather they
> work a bit harder to give a different contradiction). Clearly I've
> misunderstood what's going on - can anybody help?
>
Yes, V is consistent. That's not the question. The question is if V is
maximally consistent. If it isn't, then some q not in V with consistent
V \/ {q}. Since V \/ {q} is consistent, ~q cannot be in V. Thus V \/ {q}
and V \/ {~q} are not good, by maximally of V. Thus some finite A,B
subset V with <>(&A & q) not in V \/ {q} and <>(&B & ~q) not in V \/ {~q}.
This seems to boil down to <>q not in V \/ {q} and <>~q not in V \/ {~q}.
But isn't V \/ {<>q} good? Thus by maximality of goodness, <>q in V.
Hence V is maximally consistent. BTW, all reference to ~q aren't needed.
> Having tried to prove that V is maximal wrt consistency myself, the
> best I've managed is to show that, with q as above, we have that both
> <>q and <>1/4 q are in U. But I don't think that's enough for a
> contradiction.
>
How about that, somehow the mystrious U noted above at ** shows up here
with no apparent connection to anything being discussed. Do you proof
read your posts before commiting them to the newsgroup?
galathaea
> On Thu, 8 Jan 2009, Rotwang wrote:
> > system in question has a language L whose sentences include T and a
> > set of A atomic propositions, and are closed under the connectives 1/4,
>
> What do you mean by 1/4? Negation or ~ as rendered in ascii.
> Use plane text as characters that aren't plane text as you see
> on the top of the keys, don't show up as you expect to all readers.
your reader is not multicultural friendly
and thus incomplete
requesting others also be multicultural unfriendly
isn't really very cool...
> > &, <> and []. The axioms of the system are all substitution instances
> > of theorems of zeroth order logic (including T) and all substitution
> > instances of
>
> > [](p -> q) -> (([]p) -> ([]q)).
>
> > The rules of inference are Modus Ponens,
>
> > p p -> q
> > -------------
> > q
>
> > and necessitation,
>
> > p
> > -------------
> > []p.
>
> Are you sure that's a rule of inference for modal logic?
> Seems that would imply []p <-> p, that []p is same as p.
only if the modal logic also includes the T axiom
> > If p is a theorem we write
>
> > |- p.
>
> > More generally, let S be a set of sentences. We write
>
> > S |- p
>
> > iff there is a finite sequence of sentences p_1, p_2, ... p_n in S
> > such that
>
> > |- (p_1 & p_2 & ... p_n) -> p.
>
> > Suppose now that U is a maximal consistent set of sentences (where I
> > think that "consistent" is supposed to mean that there is no sentence
> > p such that U |- p and U |- 1/4 p, and maximal means that U u {p} is
> > inconsistent whenever p is not in U), and that <>p is in U for some p.
>
> Maximall consistent means that if p not in U, the U \/ {p} is
> inconsistent. In modal logic, I'm not so sure. You seem to be
> unsure also. Do you have an exact quote from a text to give?
there appears an additional constraint
thereExists(p) <>p e U
loosely
there is a possibility in U
> In your statement p seems to be quantified twice. Once when first
> mentioned and again at the end. Thus the meaning is not clear.
it would seem strange if p wasn't an arbitrary sentence
i'm not sure how i would parse it otherwise
> > Call a set S of sentences "good" if p in S, and whenever R is a finite
> > subset of S then <> & R is in U (here & R means r_1 & r_2 & ... r_n,
> > where r_i are the elements of R).
>
> Huh? Is there any connection between U and S? **
they are apparently collections of sentences obeying
forAll(s c S) (finite(s) -> <>(& s) e U)
> > The authors have proved that there is a maximal good set V (maximal wrt
> > being good), and that V is consistent, and are now trying to prove that
> > V is also maximal wrt consistency. Suppose otherwise: then there is some
> > q such that neither q nor 1/4 q belong to V. So neither V u {q} nor V u
> > {1/4 q} are good.
> > The authors then claim that this implies that there is a finite W c V
>
> What does the term W c V mean? Is c a binary operator? Anything wrong
> with being clear and writing W subset V. Is that what you mean?
there are many people on these groups
who assert the necessity of the use of some language or another
the city i live in
passed a law a few years ago
that required all business be done in english
it's all the same bullshit
> > such that |- & W -> q and |- & W -> 1/4 q, and use this to derive a
> > contradiction. But I don't see how this follows. In particular the fact
> > that V has already been shown to be consistent seems to imply that no
> > such W exists (and this isn't the contradiction they give - rather they
> > work a bit harder to give a different contradiction). Clearly I've
> > misunderstood what's going on - can anybody help?
>
> Yes, V is consistent. That's not the question. The question is if V is
> maximally consistent. If it isn't, then some q not in V with consistent
> V \/ {q}. Since V \/ {q} is consistent, ~q cannot be in V. Thus V \/ {q}
> and V \/ {~q} are not good, by maximally of V. Thus some finite A,B
> subset V with <>(&A & q) not in V \/ {q} and <>(&B & ~q) not in V \/ {~q}.
>
> This seems to boil down to <>q not in V \/ {q} and <>~q not in V \/ {~q}.
> But isn't V \/ {<>q} good? Thus by maximality of goodness, <>q in V.
> Hence V is maximally consistent. BTW, all reference to ~q aren't needed.
you seem to have applied the definition of good
that you had earlier seemed confused about
how confusing...
> > Having tried to prove that V is maximal wrt consistency myself, the
> > best I've managed is to show that, with q as above, we have that both
> > <>q and <>1/4 q are in U. But I don't think that's enough for a
> > contradiction.
>
> How about that, somehow the mystrious U noted above at ** shows up here
> with no apparent connection to anything being discussed. Do you proof
> read your posts before commiting them to the newsgroup?
and here you demonstrate
why the comparison with the trash in my town was relevant
i'm glad you won't read this
it let's me make my illustrations in peace...
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
Rotwang
> On Thu, 8 Jan 2009, Rotwang wrote:
> > system in question has a language L whose sentences include T and a
> > set of A atomic propositions, and are closed under the connectives 1/4,
>
> What do you mean by 1/4? Negation or ~ as rendered in ascii.
> Use plane text as characters that aren't plane text as you see
> on the top of the keys, don't show up as you expect to all readers.
Sorry. I was using the character which looks like the top right hand
corner of a rectangle, which is at the top of the key to the left of 1
on a British keyboard. I didn't realise this would be problematic,
I'll try to use ~ like you suggest from now on.
(BTW, does "plane text" have some technical meaning I don't know, or
do you mean "plain text"?)
> > &, <> and []. The axioms of the system are all substitution instances
> > of theorems of zeroth order logic (including T) and all substitution
> > instances of
>
> > [](p -> q) -> (([]p) -> ([]q)).
>
> > The rules of inference are Modus Ponens,
>
> > p p -> q
> > -------------
> > q
>
> > and necessitation,
>
> > p
> > -------------
> > []p.
>
> Are you sure that's a rule of inference for modal logic?
I have no idea whether it's /usually/ a rule. But it's in the book.
> Seems that would imply []p <-> p, that []p is same as p.
I don't think so. Compare to first order logic, where generalisation
is a rule:
P(x)
-------------
for all x (P(x))
but (P(x) -> for all x (P(x))) is nonetheless not a theorem of first
order logic.
> > If p is a theorem we write
>
> > |- p.
>
> > More generally, let S be a set of sentences. We write
>
> > S |- p
>
> > iff there is a finite sequence of sentences p_1, p_2, ... p_n in S
> > such that
>
> > |- (p_1 & p_2 & ... p_n) -> p.
>
> > Suppose now that U is a maximal consistent set of sentences (where I
> > think that "consistent" is supposed to mean that there is no sentence
> > p such that U |- p and U |- 1/4 p, and maximal means that U u {p} is
> > inconsistent whenever p is not in U), and that <>p is in U for some p.
>
> Maximall consistent means that if p not in U, the U \/ {p} is
> inconsistent. In modal logic, I'm not so sure. You seem to be
> unsure also. Do you have an exact quote from a text to give?
"The nodes Th consist of maximal, consistent subsets of K, that is,
sets U of sentences which do not allow for us to prove everything and
are maximal in this regard: if you add any other sentence, the result
is inconsistent."
The thing I wasn't sure about was what was meant by "allow us to
prove". I initially thought that U being inconsistent meant that the
theory with the sentences of U being taken as additional axioms was
inconsistent, but this made a statement which appeared later false.
But the statement in question became true with the alternative
definition given above (note that if what I wrote about p -> []p not
being a theorem is true then the deduction theorem fails for this
system, so the two notions of consistency are not equivalent).
> In your statement p seems to be quantified twice. Once when first
> mentioned and again at the end. Thus the meaning is not clear.
Oops, sorry. I meant that there is a fixed p such that <>p is in U.
For the definitions of consistency and maximal consistency, replace p
by some other letter.
> > Call a set S of sentences "good" if p in S, and whenever R is a finite
> > subset of S then <> & R is in U (here & R means r_1 & r_2 & ... r_n,
> > where r_i are the elements of R).
>
> Huh? Is there any connection between U and S? **
What? Yes. Look again:
> > Call a set S of sentences "good" if p in S, and whenever R is a finite
> > subset of S then <> & R is in ===> U <===
^
|
The definition of "good" involves U.
> > The authors have proved that there is a maximal good set V (maximal wrt
> > being good), and that V is consistent, and are now trying to prove that
> > V is also maximal wrt consistency. Suppose otherwise: then there is some
> > q such that neither q nor 1/4 q belong to V. So neither V u {q} nor V u
> > {1/4 q} are good.
> > The authors then claim that this implies that there is a finite W c V
>
> What does the term W c V mean? Is c a binary operator? Anything wrong
> with being clear and writing W subset V. Is that what you mean?
Yes, sorry.
> > such that |- & W -> q and |- & W -> 1/4 q, and use this to derive a
> > contradiction. But I don't see how this follows. In particular the fact
> > that V has already been shown to be consistent seems to imply that no
> > such W exists (and this isn't the contradiction they give - rather they
> > work a bit harder to give a different contradiction). Clearly I've
> > misunderstood what's going on - can anybody help?
>
> Yes, V is consistent. That's not the question. The question is if V is
> maximally consistent.
Yes, I know. I thought that was clear from what I wrote.
> If it isn't, then some q not in V with consistent
> V \/ {q}. Since V \/ {q} is consistent, ~q cannot be in V. Thus V \/ {q}
> and V \/ {~q} are not good, by maximally of V. Thus some finite A,B
> subset V with <>(&A & q) not in V \/ {q} and <>(&B & ~q) not in V \/ {~q}.
No, I think you've misread the definition of "good". If V \/ {q} is
not good then there is a finite A subset V with <>(&A & q) not in U.
> This seems to boil down to <>q not in V \/ {q} and <>~q not in V \/ {~q}.
> But isn't V \/ {<>q} good? Thus by maximality of goodness, <>q in V.
> Hence V is maximally consistent. BTW, all reference to ~q aren't needed.
>
> > Having tried to prove that V is maximal wrt consistency myself, the
> > best I've managed is to show that, with q as above, we have that both
> > <>q and <>1/4 q are in U. But I don't think that's enough for a
> > contradiction.
>
> How about that, somehow the mystrious [sic] U noted above at ** shows up here
> with no apparent connection to anything being discussed. Do you proof
> read your posts before commiting [sic] them to the newsgroup?
Yes. Do you?
Rotwang
> On Jan 8, 10:10 pm, William Elliot <ma...@rdrop.remove.com> wrote:
> > On Thu, 8 Jan 2009, Rotwang wrote:
>
>
> > > and necessitation,
>
> > > p
> > > -------------
> > > []p.
>
> > Are you sure that's a rule of inference for modal logic?
> > Seems that would imply []p <-> p, that []p is same as p.
>
> only if the modal logic also includes the T axiom
Really? I was under the impression that the system in question does
have T as an axiom (I needed to assume that "all substitution
instances of tautologies of classical propositional logic" included T
in order to make sense of some stuff the authors said), but that p ->
[]p is not a theorem. Do you mean that p <-> []p is a theorem iff T is
an axiom? If so, can you explain why?
Frederick Williams
>
> On Jan 8, 10:10 pm, William Elliot <ma...@rdrop.remove.com> wrote:
> > On Thu, 8 Jan 2009, Rotwang wrote:
> >
> > > and necessitation,
> >
> > > p
> > > -------------
> > > []p.
> >
> > Are you sure that's a rule of inference for modal logic?
> > Seems that would imply []p <-> p, that []p is same as p.
>
> only if the modal logic also includes the T axiom
That's not so. If one expresses MP with the diagram
p p -> q
-----------
q
then there is no reason why one shouldn't express necessitation with the
diagram
p
---
[]p .
The logic in question is called K. Including the T axiom (and nothing
else) still wouldn't make
[]p <-> p
provable.
--
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_
Frederick Williams
>
> On 9 Jan, 06:10, William Elliot <ma...@rdrop.remove.com> wrote:
> > On Thu, 8 Jan 2009, Rotwang wrote:
> > > system in question has a language L whose sentences include T and a
> > > set of A atomic propositions, and are closed under the connectives 1/4,
> >
> > What do you mean by 1/4? Negation or ~ as rendered in ascii.
> > Use plane text as characters that aren't plane text as you see
> > on the top of the keys, don't show up as you expect to all readers.
>
> Sorry. I was using the character which looks like the top right hand
> corner of a rectangle, which is at the top of the key to the left of 1
> on a British keyboard. I didn't realise this would be problematic,
> I'll try to use ~ like you suggest from now on.
>
> (BTW, does "plane text" have some technical meaning I don't know, or
> do you mean "plain text"?)
ASCII in the narrowest possible sense, so no pound sterling sign either.
> > > &, <> and []. The axioms of the system are all substitution instances
> > > of theorems of zeroth order logic (including T) and all substitution
> > > instances of
> >
> > > [](p -> q) -> (([]p) -> ([]q)).
> >
> > > The rules of inference are Modus Ponens,
> >
> > > p p -> q
> > > -------------
> > > q
> >
> > > and necessitation,
> >
> > > p
> > > -------------
> > > []p.
> >
> > Are you sure that's a rule of inference for modal logic?
>
> I have no idea whether it's /usually/ a rule. But it's in the book.
It's a rule for all normal systems (here 'normal' is a technical term).
galathaea
ugh
yeah
you are right
a nit flame has to have it's inevitable error
just as the speling flames do
i guess
H. J. Sander Bruggink
> On 9 Jan, 07:51, galathaea <galath...@gmail.com> wrote:
>> On Jan 8, 10:10 pm, William Elliot <ma...@rdrop.remove.com> wrote:
>>> On Thu, 8 Jan 2009, Rotwang wrote:
>> [...]
>>
>>>> and necessitation,
>>>> p
>>>> -------------
>>>> []p.
>>> Are you sure that's a rule of inference for modal logic?
>>> Seems that would imply []p <-> p, that []p is same as p.
No, it doesn't imply that. It says that |- p implies |- []p, not
that |- p -> []p. In normal first-order logic the above would
be equivalent, but in modal logical, they're not.
>> only if the modal logic also includes the T axiom
>
> Really? I was under the impression that the system in question does
> have T as an axiom (I needed to assume that "all substitution
> instances of tautologies of classical propositional logic" included T
> in order to make sense of some stuff the authors said), but that p ->
> []p is not a theorem. Do you mean that p <-> []p is a theorem iff T is
> an axiom? If so, can you explain why?
T is the axiom []p -> p. This is is not a tautology of classical
propositional logic, because q -> p is not.
groente
-- Sander
H. J. Sander Bruggink
>> Suppose now that U is a maximal consistent set of sentences (where I
>> think that "consistent" is supposed to mean that there is no sentence
>> p such that U |- p and U |- 殆, and maximal means that U u {p} is
>> inconsistent whenever p is not in U), and that <>p is in U for some p.
>> Call a set S of sentences "good" if p in S, and whenever R is a finite
>> subset of S then <> & R is in U (here & R means r_1 & r_2 & ... r_n,
>> where r_i are the elements of R). The authors have proved that there
>> is a maximal good set V (maximal wrt being good), and that V is
>> consistent, and are now trying to prove that V is also maximal wrt
>> consistency. Suppose otherwise: then there is some q such that neither
>> q nor 段 belong to V. So neither V u {q} nor V u {段} are good. The
>> authors then claim that this implies that there is a finite W c V such
>> that |- & W -> q and |- & W -> 段, and use this to derive a
>> contradiction. But I don't see how this follows. In particular the
>> fact that V has already been shown to be consistent seems to imply
>> that no such W exists (and this isn't the contradiction they give -
>> rather they work a bit harder to give a different contradiction).
>> Clearly I've misunderstood what's going on - can anybody help?
Since V is maximally good, V u {q} is not good. By definition, this
means that there is a finite subset W of V u {q} such that <> &W is
not in U.
Now suppose W does not contain q. Then W would be a subset of V and
then, by definition, <> &W would be in U, which is a contradiction.
So, W must contain q.
That means that &W is a big conjunction containing p. Then it must
hold that |- W -> q. (For example, if W = a & q, then clearly
|- a & q -> q.)
The same holds for ~p.
I hope this helps.
groente
-- Sander
Rotwang
Ah, I see. I was taking T to mean the trivial statement, i.e. the
statement which satisfies x |= T for any set x, hence my confusion.
Thanks.
Rotwang
> Rotwang wrote:
> >> Suppose now that U is a maximal consistent set of sentences (where I
> >> think that "consistent" is supposed to mean that there is no sentence
> >> inconsistent whenever p is not in U), and that <>p is in U for some p.
> >> Call a set S of sentences "good" if p in S, and whenever R is a finite
> >> subset of S then <> & R is in U (here & R means r_1 & r_2 & ... r_n,
> >> where r_i are the elements of R). The authors have proved that there
> >> is a maximal good set V (maximal wrt being good), and that V is
> >> consistent, and are now trying to prove that V is also maximal wrt
> >> consistency. Suppose otherwise: then there is some q such that neither
> >> authors then claim that this implies that there is a finite W c V such
> >> contradiction. But I don't see how this follows. In particular the
> >> fact that V has already been shown to be consistent seems to imply
> >> that no such W exists (and this isn't the contradiction they give -
> >> rather they work a bit harder to give a different contradiction).
> >> Clearly I've misunderstood what's going on - can anybody help?
Thanks for your reply, but I'm afraid I don't follow. Questions
follow.
>
> Since V is maximally good, V u {q} is not good. By definition, this
> means that there is a finite subset W of V u {q} such that <> &W is
> not in U.
>
> Now suppose W does not contain q. Then W would be a subset of V and
> then, by definition, <> &W would be in U, which is a contradiction.
> So, W must contain q.
I'm with you so far.
>
> That means that &W is a big conjunction containing p.
Do you mean q here?
> Then it must
> hold that |- W -> q. (For example, if W = a & q, then clearly
> |- a & q -> q.)
>
> The same holds for ~p.
Do you mean ~q here? If not then I don't follow. On the other hand, if
so then this shows that there are sets W subset V u {q} and X subset V
u {~q} such that |- &W -> q and |- &X -> ~q. But I don't see how to
get a contradiction from here - would you mind going into more detail?
Sander Bruggink
> On 9 Jan, 17:12, "H. J. Sander Bruggink" <brugg...@uni-due.de> wrote:
> > Since V is maximally good, V u {q} is not good. By definition, this
> > means that there is a finite subset W of V u {q} such that <> &W is
> > not in U.
>
> > Now suppose W does not contain q. Then W would be a subset of V and
> > then, by definition, <> &W would be in U, which is a contradiction.
> > So, W must contain q.
>
> I'm with you so far.
>
> > That means that &W is a big conjunction containing p.
>
> Do you mean q here?
Yes, I meant q, but I see I proved the wrong thing, because W is not a
subset of V.
So W contains q. Let W' be the rest of W, that is W = W' u {q}.
We know that <> &W' is in U, but <> &W is not. Since U is
maximally consistent, this can only be the case, if &W' -> ~q.
> > The same holds for ~p.
>
> Do you mean ~q here?
Yes, I meant ~q there.
> If not then I don't follow. On the other hand, if
> so then this shows that there are sets W subset V u {q} and X subset V
> u {~q} such that |- &W -> q and |- &X -> ~q. But I don't see how to
> get a contradiction from here - would you mind going into more detail?
We have a W' subset V such that &W' -> ~q and a set X' such that
&X' -> q. Now you can take the union of these two sets to get the
set you were after.
I realize the names of the sets are a bit different than the ones you
used, but the idea should be clear.
groente
-- Sander
Rotwang
> On Jan 9, 8:00 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
> > On 9 Jan, 17:12, "H. J. Sander Bruggink" <brugg...@uni-due.de> wrote:
> > > Since V is maximally good, V u {q} is not good. By definition, this
> > > means that there is a finite subset W of V u {q} such that <> &W is
> > > not in U.
>
> > > Now suppose W does not contain q. Then W would be a subset of V and
> > > then, by definition, <> &W would be in U, which is a contradiction.
> > > So, W must contain q.
>
> > I'm with you so far.
>
> > > That means that &W is a big conjunction containing p.
>
> > Do you mean q here?
>
> Yes, I meant q, but I see I proved the wrong thing, because W is not a
> subset of V.
>
> So W contains q. Let W' be the rest of W, that is W = W' u {q}.
> We know that <> &W' is in U, but <> &W is not.
Right.
> Since U is
> maximally consistent, this can only be the case, if &W' -> ~q.
I'm afraid I don't see how you conclude this. Sorry if I'm being slow
- like I say this is my first encounter with modal logic. Can you
expand on this part of the argument?
>
> > > The same holds for ~p.
>
> > Do you mean ~q here?
>
> Yes, I meant ~q there.
>
> > If not then I don't follow. On the other hand, if
> > so then this shows that there are sets W subset V u {q} and X subset V
> > u {~q} such that |- &W -> q and |- &X -> ~q. But I don't see how to
> > get a contradiction from here - would you mind going into more detail?
>
> We have a W' subset V such that &W' -> ~q and a set X' such that
> &X' -> q. Now you can take the union of these two sets to get the
> set you were after.
>
> I realize the names of the sets are a bit different than the ones you
> used, but the idea should be clear.
Yes, I see how to get a contradiction from here, assuming the above
step.