On 02 Apr 2014, at 23:20, LizR wrote:
On 3 April 2014 04:37, Bruno Marchal <[email protected]> wrote:
Suppose R is not transitive, so for all beta (alpha R beta) and
there are some gamma such that [(beta R gamma) and ~(alpha R gamma)].
I cannot parse that sentence, I guess some word are missing. R is
not transitive means that there exist alpha, beta and gamma, such that
alpha R beta, and beta R gamma, and ~(alpha R gamma). I will guess
that this is what you meant.
That's what I took it to mean. (I didn't realise that wasn't what it
said!)
As a math teacher I am aware that when a student cannot solve a
problem, it is very often due to their inability to read a text
literally.
To progress in math you have to try to be dumber, not cleverer.
Especially in logic.
OK Liz? Others? Feel free to ask definitions or explanations.
Yes, at least at the point where I think very hard about each one,
they all seem to make sense.
Good, but you might need to train yourself so that "seems" becomes
"pretty sure".
The next one is important, as it plays a role in the 'derivation of
physics'.
> (W, R) respects A -> []<>A if and only R is symmetrical,
R symmetrical means that if (alpha R beta) then (beta R alpha).
Yes, for all alpha and beta in W.
Suppose A is true in alpha; then <>A is true in beta (by symmetry
of R) and this holds for all alpha and beta so []<>A in alpha.
And so A -> []<>A is true in alpha. (Here we are using the
deduction rule in the CPL context, which is valid. Later we will see
it is not valid in the modal context).
Suppose R is not symmetrical, so there is a pair of worlds (alpha R
beta) and ~(beta R alpha). So consider V such that A=t in alpha
and A=f in all worlds gamma such that (beta R gamma) then ~<>A in
beta. So it would be false that []<>A in alpha.
Liz told me this already! OK.
Phew.
> (W,R) respects []A -> <>A if and only if R is ideal,
R is ideal, means that for every alpha there is a beta such that
(alpha R beta). Suppose []A is true in alpha, then A must be true
in every world beta (alpha R beta) and there is a least on such
beta, so <>A is true in alpha.
OK.
Suppose R is not ideal, then there is a cul-de-sac alpha. For
alpha []A is vacously true for all A, but <>A is false so []A-><>A
is false.
Yes, all cul-de-sac world are counterexample of []A -> <>A. In the
Kripke semantics, they are counterexamples of <>#, with # put for
any proposition.
> (W, R) respects <>A -> ~[]<>A if and only if R is realist.
R is realist means that for every world alpha there is a world beta
such that (alpha R beta) and beta is cul-de-sac.
For every *transitory* world alpha. OK. The cul-de-sac world are
still world!
Suppose A is true in beta, then <>A is true in alpha but <>A=f in
beta so []<>A cannot be true in alpha. Hence <>A->~[]<>A in alpha
where alpha is any non cul-de-sac world. Then consider a cul-de-
sac world like beta; <>A is always false in beta so <>A->X is true
in beta for any X, including ~[]<>A.
OK. Nice.
So you proved that R is realist implies that (W, R) respects <>A ->
~[]<>A.
But you have still not prove that if R is *not* realist, (W,R) does
not respect <>A -> ~[]<>A (unlike all other cases). OK?
You proved: "(W, R) realist" implies "respects <>A -> ~[]<>A", but
not yet the converse, that "respects <>A -> ~[]<>A" implies " (W, R)
realist".
I let you search, and might justify this (with pre-warning to avoid
spoiling!).
And what about the euclidian multiverse? May be you did them?
R is euclidian, or euclidean, if (aRb and aRc) implies bRc, for all
a, b and c in W. (I use "a" for the greek alpha!)
Proposition: (W,R) respects <>A -> []<>A iff R is euclidian.
Hmm. I'll think about that later.
OK. The time you are using to learn is not important, unless you ...
forget the work already done. For the long term memorizing, it is
better to revise 2 minutes everyday, instead of learning day and night
just before the exams :)
Bruno
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