On 20 Jan 2014, at 23:47, LizR wrote:
On 21 January 2014 08:38, Bruno Marchal <marc...@ulb.ac.be> wrote:
If you remember Cantor, you see that if we take all variables into
account, the multiverse is already a continuum. OK? A world is
defined by a infinite sequence like "true, false, false, true, true,
true, ..." corresponding to p, q, r, p1, q1, r1, p2, q2, ...
I assume it's a continuum, rather than a countable infinity because
if it was countable we could list all the worlds, but of course we
can diagonalise the list by changing each truth value.
Very good.
(Those who does not get this can ask for more explanations).
On 21 Jan 2014, at 01:32, LizR wrote:
On 21 January 2014 08:38, Bruno Marchal <marc...@ulb.ac.be> wrote:
Are the following laws? I don't put the last outer parenthesis for
reason of readability.
p -> p
This is a law because p -> q is equivalent to (~p V q) and (p V ~p)
must be (true OR false), or (false OR true) which are both true
(p & q) -> p
using (~p V q) gives (~(p & q) V q) ... using 0 and 1 for false and
true ... (0,0), (0,1) and (1,0) give 1, (1,1) gives 1 ... so this is
true. So it is a law. I think.
(p & q) -> q
Hmm. (~(p & q) V q) is ... the same as above.
p -> (p V q)
(~p V (p V q)) must be true because of the p V ~p that's in there
(as per the first one)
q -> (p V q)
Is the same...hm, these are all laws (apparently). I feel as though
I'm probably missing something and getting this all wrong. Have I
misunderstood something ?
No, it is all good, Liz!
What about:
(p V q) -> p
and
p -> (p & q)
What about (still in CPL) the question:
is (p & q) -> r equivalent with p -> (q -> r)
Oh! You did not answer:
((COLD & WET) -> ICE) -> ((COLD -> ICE) V (WET -> ICE))
So what? Afraid of the logician's trick? Or of the logician's madness?
Try this one if you are afraid to be influenced by your intuition
aboutCOLD, WET and ICE:
((p & q) -> r) -> ((p -> r) V (q -> r))
Is that a law?
And what about the modal []p -> p ? What about the []p -> [][]p, and
<>p -> []<>p ? Is that true in all worlds?
Let me an answer the first one: []p -> p. The difficulty is that we
can't use the truth table, (can you see why) but we have the meaning
of "[]p". Indeed it means that p is true in all world.
Now, p itself is either true in all worlds, or it is not true in all
worlds. Note that p -> p is true in all world (as you have shown
above, it is (~p V p), so in each world each p is either true or false.
If p is true in all worlds, then p is a law. But if p is true in all
world, any "A -> p" will be true too, given that for making "A -> p"
false, you need p false (truth is implied by anything, in CPL). So if
p itself is a law, []p -> is a law. For example (p->p) is a law, so []
(p->p) -> (p->p) for example.
But what if p is not a law? then ~[]p is true, and has to be true in
all worlds. With this simple semantic of Leibniz, []p really simply
means that p is true in all world, that is automatically true in all
world. If p is not a law, ~[]p is true, and, as I said, this has to be
true in all world (in all world we have that p is not a law).
So []p is false in all worlds. But false -> anything in CPL. OK? So
[]p -> p is always satisfied in that case too.
So, no matter what, p being a law or not, in that Leibnizian universe:
[]p -> p *is* a law.
In Leibniz semantic, you have just a collection of worlds. If []p is
true, it entails that p is true in all worlds, so []p is true in all
world too.
Can you try to reason for []p -> [][]p, and <>p -> []<>p ? What about
p -> []p ? What about this one:
([]p & [](p -> q)) -> []q ?
Ask any question, up to be able to find the solution. tell me where
you are stuck, in case you are stuck. I might go too much quickly, you
have to speed me down, by questioning!
It may seem astonishing, but with the simple Leibnizian semantic, we
can answer all those questions.
With Kripke semantics, the multiverse will get more structure. But in
Leibniz, all worlds are completely independent, so if p is a law, the
fact that it is a law is itself a law, and []p will be true in all
worlds, and be a law itself. Indeed if []p was not a law, there would
be a world where ~p is true, and p would not be a law, OK?
Take those questions as puzzle, or delicious torture :)
Bruno
http://iridia.ulb.ac.be/~marchal/
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an
email to everything-list+unsubscr...@googlegroups.com.
To post to this group, send email to everything-list@googlegroups.com.
Visit this group at http://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/groups/opt_out.
