Modal logic: a simple question
Alberto M. Scattolo
I'm studying modal logic and I've a question that probably is simple:
I need to prove that □p --> ◊p, also known as axiom D, is not valid in
the class of all frames?
If you can read correctly all chars it is: box p implies diamond p.
I belived that axioms are always valid... but it seems I'm wrong.
Thanks a lot in advance.
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Frederick Williams
>
> Hi everybody!
>
> I'm studying modal logic and I've a question that probably is simple:
>
> I need to prove that □p --> ◊p, also known as axiom D, is not valid in
> the class of all frames?
>
> If you can read correctly all chars it is: box p implies diamond p.
Write it as []p -> <>p or Lp -> Mp .
> I belived that axioms are always valid... but it seems I'm wrong.
D is an axiom of the logic KD (and true in every KD-mode) but it is not
true in every model.
--
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
> D is an axiom of the logic KD (and true in every KD-mode) but it is not
> true in every model.
Sorry, I meant KD-model not KD-mode.
Aatu Koskensilta
> I need to prove that []p --> <>p, also known as axiom D, is not valid in
> the class of all frames?
(I have replaced the special characters in the above with ASCII
approximations.) Can you think what properties a world w and the
worlds immediately accessible from it should have in order for D to be
false in w? Recall the truth conditions for []p and <>p, and that an
implication is false if and only if its antecedent is true but its
consequent is false.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Alberto M. Scattolo
Now, how to demostrate it? Trying to explying []p -> <>p I would says
that:
if, for all world w1, with wRw1, M,w1 |= p, then, some world w2, with
wRw2, M,w2 |= p
Here we have two cases:
1) M,w |= []p is false, then []p -> <>p is satisfiable
2) M,w |= []p is true. If it is true M,w |= []p, when M,w |= <>p can
be false? If we assume it true fo every world... it can't be false for
just some world.
This would lead me to say that it's valid but we already said it's
not.
What am i missing?
Which are all possible models?
Thanks a lot again!
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Aatu Koskensilta
> Here we have two cases:
> 1) M,w |= []p is false, then []p -> <>p is satisfiable
> 2) M,w |= []p is true. If it is true M,w |= []p, when M,w |= <>p can
> be false?
Consider the statements
(a) All real numbers satisfying P(r) also satisfy Q(r).
(b) There is a real number satisfying P(r) and Q(r).
Can you think of properties P and Q of reals such that (a) is true but
(b) is false? The situation with (1) and (2) above is analogous.
> What am i missing?
Does the fact that, for any world w', if w' is accessible from w, then
w' |= p, in itself tell us anything about whether there exists any w'
accessible from w? Recall that for any w', by the truth conditions for
implications the statement "if w' is accessible from w, then w' |= p"
is false only if the antecedent, "w' is accessible from w", is true
but the consequent, "w' |= p" is not; and thus that "for any world w',
if w' is accessible from w, then w' |= p" is false only if there is
some world w' that is accessible from w in which p is false.
Alberto M. Scattolo
On Feb 3, 4:45 pm, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> Consider the statements
>
> (a) All real numbers satisfying P(r) also satisfy Q(r).
> (b) There is a real number satisfying P(r) and Q(r).
>
> Can you think of properties P and Q of reals such that (a) is true but
> (b) is false? The situation with (1) and (2) above is analogous.
Unfortunately, nothing comes to my mind :(
> Does the fact that, for any world w', if w' is accessible from w, then
> w' |= p, in itself tell us anything about whether there exists any w'
> accessible from w? Recall that for any w', by the truth conditions for
> implications the statement "if w' is accessible from w, then w' |= p"
> is false only if the antecedent, "w' is accessible from w", is true
> but the consequent, "w' |= p" is not; and thus that "for any world w',
> if w' is accessible from w, then w' |= p" is false only if there is
> some world w' that is accessible from w in which p is false.
Ok, I completely agree with this. You say something interesting:
the sentence
if w' is accessible from w then w' |= p
actually says nothing about w' existence. But if no w' exists, can I
say that []p is true?
Of course if I can say that []p is true even when there's no w', then
the implication is false since accessible worlds can not be |= p.
Thanks again, it's very important for me :)
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Aatu Koskensilta
> You say something interesting: the sentence
>
> if w' is accessible from w then w' |= p
>
> actually says nothing about w' existence. But if no w' exists, can I
> say that []p is true?
Why not? By definition, []p is true in w if and only if []p is not
false in w, or, in other words, if there are no worlds accessible from
w in which p is false. If there are no worlds accessible from w at
all, there certainly aren't any in which p is false (or true for that
matter).
Alberto M. Scattolo
Aatu thanks a lot for you help, I really apriciate it!
Have a nice day!
I'll be back if I have other doubts :)
Bye :D
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Ken Pledger
<b4906342-967a-42d8...@g3g2000pre.googlegroups.com>,
"Alberto M. Scattolo" <thedark...@gmail.com> wrote:
> ....
> I need to prove that ещp --> вp, also known as axiom D, is not valid in
> the class of all frames?
>
> If you can read correctly all chars it is: box p implies diamond p.
> I belived that axioms are always valid... but it seems I'm wrong....
An easier notation to write is Lp -> Mp. That can easily be
transformed into M(p -> p), i.e. M followed by a formula which comes
out true in every world in every model So if a world can see any world
at all then your formula must be true in it. Is that easier to think
about?
Ken Pledger.
Alberto M. Scattolo
I think that may be it's more intuitive to me to read Lp as:
all accessible nodes |=p which means that there are no accessible
nodes such that |= ¬p.
Knowing that L does not implies the existence of an accessible nodes
is important to me not only for this case but also in other exercises,
and it also helped me understanding that, on the contrary, M says that
actually a node exists.
I was missing this understanding from lectures notes. :)
Thanks again to you and to everybody.
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