[FOM] Is axiom R sound on a topological space?
twilson at csufresno.edu
Fri Nov 8 13:50:34 EST 2002
On Fri, 8 Nov 2002, abuchan at mail.unomaha.edu wrote:
> It is now popular to interpret the box operator in modal logic S4 as the
> interior operator on a topological space and the diamond as the closure
> following Tarski and McKinsey's paper 'The Algebra of Topology'. If we
> pick the right axioms for S4 that correspond nicely with Kuratowski's
> axioms for closure and interior then it is a relatively exercise to show
> that S4 is sound. However, it is not at all clear that the
> axiom (R):= BOX(p->q)->BOX(p)->BOX(q) is sound. Does anyone see the proof
> for this?
Write subset as <=, intersection as /\, union as \/, define
A -> B = int((X - A) \/ B),
and note that int (is monotone and) preserves finite intersections:
int(A /\ B) = int(A) /\ int(B).
Since all intuitionistic identities hold under this interpretation,
int(A->B) -> (int(A) -> int(B)) = (int(A->B) /\ int(A)) -> int(B)
= int((A->B) /\ A) -> int(B)
= int(A /\ B) -> int(B)
Todd Wilson A smile is not an individual
Computer Science Department product; it is a co-product.
California State University, Fresno -- Thich Nhat Hanh
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