On 24 Jan 2014, at 21:52, LizR wrote:
On 24 January 2014 23:05, Bruno Marchal <[email protected]> wrote:
On 24 Jan 2014, at 00:01, LizR wrote:
On 24 January 2014 00:33, Bruno Marchal <[email protected]> wrote:
(Later, we will stop asking that all worlds (in the sense given)
belongs in the multiverse. We can decide to suppress all worlds in
the multiverse in which p is true. And keep Leibniz semantics in
that new sort of multiverse (meaning that []x is true = x is true
in all worlds of that multiverse).
Question: how to add one word in your justification above, so that
"p -> []<>p" is still justified as a law. Which word?)
I don't know. I'm confused.
I give then answer.
Well I repeat the question first. With the definition I gave of
"world" the Multiverse has to contain
1) all possible valuation (assignment of 0 or 1) of the letters (=
propositional variables).
2) cannot contain two worlds with the same valuation.
But we can generalize a bit the semantics of Leibniz, by abandoning
1) and 2).
Then your argument that p-> []<>p, which was
"it's true in all worlds that p is true in at least one world."
becomes
if p is true (in this world, say) then it's true in all worlds that
p is true in at least one world.
You need just use a conditional (if). The word asked was "if".
OK?
OK. I think I see. p becomes "if p is true" rather than "p is true"
Yes.
Rereading a previews post I ask myself if this is well understood.
You said that we cannot infer anything from Alicia song as we don't
know if his theory/song is true.
But the whole point of logic is in the art of deriving and reasoning
without ever knowing if a premise is true or not. Indeed, we even want
to reason independetly of any interpretation (of the atoical
propositions).
That error is done by those who believe that I defend the truth of
comp, which I never do.
In fact we never know if a theory is true (cf Popper). That is why we
do theories. We can prove A -> B, without having any clues if A is
false (in which case A -> B is trivial), or A is true.
I will come back on this. It is crucially important.
A good example is Riemann Hypothesis (RH). We don't know if it is
true, but thousand of papers study its consequence.
If later we prove the RH, we will get a bunch of beautiful new theorem.
If we discover that RH leads to a contradiction, then we refute RH,
and lost all those theorems, but not necessarily the insight present
in some of the proofs.
Just to be sure: ~[]<>p is the negation of []<>p. But of course "<>p
-> ~[]<>p" is not the negation of
"<>p -> []<>p". OK?
By the way what is the negation of (p -> q)? is it (~p -> ~q)? Or is
it (~q -> ~p)?
~(p -> q) ?
Or if ~
Or even if 1...
I guess p -> q is 'the truth of p implies the truth of q" or "if p
is true then q is true". If p is false, q could be true or false.
It's equivalent to (~p V q) which means if p is false, the result is
true regardless of q, if p is true, the result is q.
The negation of that would be if p is false, q is true, wouldn't it?
So (~p -> q)
?
Also it would be ~(~p V q) which maybe comes out to - at a guess (p
V ~q). Or does it?
ok I need a truth table!
There is no shame to come back to a truth table. May be some day I can
explain you the tableau method, which is handy to verify CPL
statements, and can be extended to modal logic, but this is not really
part of the "course" ...
p,q=00 = 0, 01=0, 10=1, 11=0. Call that 0010 (That looks
familiar...) So the negation would have to be 1101.
p V ~q gives 00=1 01=0 10=1 11=1 so not quite the same! So you can't
"multiply through" by ~ (somehow that doesn't surprise me)
Let's try ~p -> q, that's (~~p V q) or p V q which is true except
for 00. SO not that either!
OK let's stop trying to work out the answer and use the hint you
gave. Is it (~p -> ~q)? Or is it (~q -> ~p)?
The first one is (p V ~q) which gives 00=1 01=0 10=1 11=1 or 1011,
which isn't the negation I got above 1101
The second gives (q V ~p) which gives 00=1 01=1 10=0 11=1 which is
1101, so that's the right answer.
(~q -> ~p) "if q is false that implies p is false"
The negation of (p -> q) = ~(p -> q) = ~(~p V q) = ~~p & ~q = p & ~q.
That's all. It describes the only line where (p -> q) is false. p must
be false and q true.
What you show above is that (p -> q) is equivalent with (~q -> ~p).
It is the correct answer to a wrong question. My fault. I am not sure
why I add that "negation". Sorry.
There was a trap, but I did not intent it.
Are you convinced that in a Leibnizian multiverse (even with our
generalized definition) we have the following laws (with their usual
name):
[]p -> p (T)
[]p -> [][]p (4)
[](p -> q) ->. ([]p -> []q) (k)
[]p -> <>p (D)
p -> []<>p (B)
<>p -> []<>p (5)
and that we don't have (as law)
p -> []p (Triv)
<>p -> ~[]<>p (g)
OK, I believe I proved those (but now I need to read them as English
to understand them again, I am no good with the symbols).
Yes. OK. You can read them neutrally, for example, for <> p -> [] <>
p, you can read this
(diamond p) implies (box (diamond p)), or
IF diamond p, then box (diamond p).
That is useful when you don't want to interpret the modalities. When
you are in a purely syntactic mode of reflexion.
Now, in Leibniz semantic, the reading "necessary" and "possible" makes
a lot of sense. it is really the intuition of possible and necessary
in some strong "absolute sense", and so the same formula can be read:
if p is possible, then p is necessary possible, which makes sense both
intuitively, and in Leibniz semantics.
All the formula above (well, except the last two which are not laws)
are considered intuitively valid for that strong notion of necessary
(true in all worlds) and possible (true in at least one world).
Of course, if some formula are not laws in a multiverse, it remains
possible that such formula are true in *some* world of the
multiverse. Can you build a little multiverse in which those last
two formula are true in some world? or is the negation of such
formula laws?
OK, p -> []p
The negation is ~[]p -> ~p
"if it's not true that p is true in all worlds, that implies p is
false in this world" (?)
That isn't a law. Nor is the truth of p in this world proof of it
being true in all worlds. I can make a multiverse in which p is
always true, say one world where p is true. In that "multiverse" p -
> []p
OK. I guess you meant "p is not always true".
As "p -> []p" contains only one propositional variable, p. A precise
multiverse in which p->[]p is true in one world and false in another
would be simply a multiverse with a world where p is true and a world
were p is false. In the world where p is false, we have []p is false
too, and so p->[]p is true in that world. But in the other world,
where p is true, []p is still false (as not being true in all worlds),
so p -> []p is false in that second world, and true in the first one.
OK?
Careful, a proposition can be such that neither its or its negation
are laws.
We have p V ~q is a law, but []p V []~p is not a law! (thing of a
multiverse having worlds in which p is true, and worlds in which p is
false).
<>p -> ~[]<>p
"if p is true in some world that implies that in not all worlds, p
is true in some world" - which isn't true.
Which is not a law. But you are supposed to find a world in which it
is still true. Is that possible?
With our new leibnizian multiverse, it is. Imagine that p is false in
all worlds. Then <>p is false in all worlds, but then <>p -> # (with
# being anything you want) is true, as the false implies any
proposition (f -> p) is a law!
And so (<>p -> ~[]<>p) is true in all worlds of a multiverse where p
is false everywhere. OK?
With the older leibnizian multi-universe, as we take *all* valuation
as world, there is a world with p true, and thus <>p is always true,
but then []<>p too, and so (<>p -> ~[]<>p) is never satisfied, in any
world, of such multiverse. OK?
The negation is []<>p -> <>p which is just []p -> p, which is "T"
hence a law.
I think! :-)
Not sure why you take the negation.
Also, the negation of (p->q) is (p & ~q), thus the negation of <>p ->
~[]<>p is <>p & []<>p.
I am afraid I have inadvertently misleaded you with my "typo" error
above. By the truth table of (p->q), the negation is a conjunction.
But here you make also a genuine error, not due to my bad teaching and
typo!
You cannot say that []<>p -> <>p is just []p -> p.
It is correct that if []p -> p is valid, (true in all world) then
[]<>p -> <>p is too. But from []<>p -> <>p, you cannot generalize into
[]p -> p.
Look Liz, I think that sometimes I wrote the posts too quickly.
I give you holiday, and I will send a clearer post for the
continuation, when I got more time to be sure that I will not mislead
you by typo error.
Just try to read the formula in english, with [] = necessary and <>
possible, and to intuit that the formula above are indeed laws (true
in all worlds) in all Leibnizian multiverse.
You are doing good job, unlike me. Sorry!
Good week-end,
Bruno
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