In article
763503bd-a2dc-41a2-87e9-6cae2a3cb... at (no spam) z66g2000hsc.googlegroups.com>,

 translogi <wilem... at (no spam) googlemail.com> wrote:
....
the riddle is in Modal logic S3 is

"
[][](p -> p) v <><>q is a theorem
but
[][](p -> p) nor
 <><>q
is a theorem
"

      I'm not sure why you call it a "riddle".  It shows why S3 (like
S1, S2 and other systems) is often called "unreasonable" in the sense of
Hallden.







The solution I came to is that the theorem is an Law of the excluded
middle theorem

starting with

A v ~A
taking A to be [][]T

(T being any tautaulogy)

We get
[][]T v ~[][] T

converting ~[][] to <><>~
we get
 [][]T v <><>~T

because every tautiology is equivalent to any (other or the same)
tautology we can replace the first one by p ->p and the second one by
q v -q

we get
[][](p->p) v <><>~(q v ~q)

~(q v ~q) is equivalent to ~q & q
we get
[][](p->p) v <><>(~q & q)

(~q & q) implies <><>q
so we get

[][](p->p) v <><>q

 QED

Is this proof correct?

      Yes, even in S1.  It's pretty similar to the proof in Hughes &
Cresswell's first book "An Introduction to Modal Logic" pp. 268-9.  Have
you seen some essentially different proof somewhere?

            Ken Pledger.- Hide quoted text -

- Show quoted text -