On 02 Apr 2014, at 23:15, LizR wrote:
As instructed I will have a look at Brent's proofs and see if I
follow them, and agree...
On 2 April 2014 15:45, meekerdb <[email protected]> wrote:
On 4/1/2014 7:40 AM, Bruno Marchal wrote:
BTW, are you OK in the math thread? Are you OK, like Liz apparently,
that the Kripke frame (W,R) respects A -> []<>A iff R is symmetrical?
Should I give the proof of the fact that the Kripke frame (W,R)
respects []A -> [][]A iff R is a transitive?
Bruno
Here's the ones I've done so far. One more to go. Hold off on that
proof (or put a warning in the subject line so I can avoid reading
it).
Brent
> *******************
> Show that
>
> (W, R) respects []A -> A if and only if R is reflexive,
R is reflexive implies (alpha R alpha) for all alpha. []A in alpha
implies A is true in all beta where (alpha R beta), which includes
the case beta=alpha. So R is reflexive implies (W,R) respects []A->A.
I like more words, but I think I follow that and it comes out right.
Assume R is not reflexive. Then there exists at least one world
beta such that (alpha R beta) and ~(beta R alpha). Consider a
valuation such that p=f in alpha and p=t in all beta. Then []p is
true in alpha but p is false so []A->A is false in alpha for some
A. R not reflexive implies []A->A is not respected for all alpha
and all valuations.
Yes that seems right, too. Brent obviously has a far more logical
mind than I do, but I guess I already knew that.
> (W, R) respects []A -> [][]A if and only R is transitive,
R is transitive means that for all beta such that (alpha R beta) and
all gamma such that (beta R gamma), (alpha R gamma). So every []A
implies A=t in all beta and also A=t in all gamma. But A=t in all
gamma means []A is true in beta, which in turn means [][]A is true
in alpha. So R is transitive implies (W,R) respects []A->[][]A.
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)]. Let A=t in beta, A=f in gamma. Then []A is true in alpha
but []A isn't true in beta, so [][]A isn't true in alpha. So (W, R)
respects []A -> [][]A implies R is transitive.
Yes, again, I eventually managed to follow that. You make it seem so
easy.
> (W, R) respects A -> []<>A if and only R is symmetrical,
R symmetrical means that if (alpha R beta) then (beta R alpha).
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.
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.
Again I an overawed.
> (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.
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.
> (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. 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.
I think my brain is starting to melt down, I can't work out if that
proves "if and only if" ?
By the way why "realist" ?
By lack of imagination of my part. The idea that is that "we can die
at each instant (in each "world")" looks realist. " <>A -> ~[]<>A i
the main axiom of the "smallest theory of life/intelligence".
OK, Liz, but sometimes you say "it seems correct". You should perhaps
try to convince your son, or Charles, to develop the confidence.
Bruno
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