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Oops, I was a little bit too fast:
not u == ~p+ & ~p-
So we probably need the u costant as well.
Or we can do the following trick. Instead that
we derive from theory T, a formula A without
u, we simply add the following axiom:
~[]u & <>u (X)
Then we can derive from T,X a formula B
that contains u, and u will behave
correctly.
Or put it the other way, the following formula
X->B
where B might contain u, and X
is defined as above, behaves as u would
have the value U assigned.
Bye
Jan Burse wrote:
Hi
Confutus wrote:
> It's [](_ -> _} and [](_< -> _} applied to
> the Lukasiewicz conditional that seem to have
> gone unappreciated.
What are the curly braces above ("}"), and the
plonked less than sign ("<") ?
> Granted, but without a well-behaved conditional
> this is only a fragment of a full-featured logic.
BTW:
(A->B)+ = A- v (A+ & B+) v (~A- & ~A+ & ~B-)
(A->B)- = A+ & B-
Which can easily be read off from your 3-valued
table, thus if you had the constant u in your
system L3 you could represent A->B by the other
connectives. Namely by combining (A->B)+ and (A->B)-
as I suggested in my other post.
But you dont need the constant u in your system.
You can use an arbitrary propositional variable p,
then we have:
u == ~p+ & ~p-
Or when we translate it back to your "modal" operators,
we then have:
u == ~[]p & <>p
further notice that ~[]~A == <>A. Thus your whole
system could be reduced to &, v, ~ and [].
Bye
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