How to prove that the axiom H is valid in the following frame

Question

To prove incompleteness of KH, I have to prove that the axiom H is valid on the following frame:

Axiom H goes as follows:

□(□p↔p)→□p

I don't know how to prove this, but here's one idea that's half complete. To prove the formula is valid on this frame we need to prove that we can't falsify it. We can separate this frame into two: any world is either world 0, or it is a world that sees world 0. World 0 does not see any other world, so □p is true in 0. Since □p is true, the entirety of the implication must be true, and so the axiom is true in 0. For any world that belongs to N, let's try to falsify the axiom. If the axiom were false, then in world n we'd have:

□(□p↔p) ∧ ¬□p

Since the world n sees 0, □p↔p must be true in 0. But since □p is true in 0 (since it sees no worlds), p must be true in 0 too. Since 1 sees only 0 and p is true in 0, □p must be true in 1. Since n sees 1, □p↔p is true in 1, so p is true in 1. We can continue this upwards all the way to n+1, so therefore, it can't be the case that ¬□p. Same can be shown for any n that belongs to N, so the axiom is not falsifiable for if n belongs to N.

Here however I get stuck. I need to prove the same for if n belongs to N*. However, I no longer have the luxury of the upwards climb because there's infinitely many worlds in between n and m. If I attempt to falsify H at some world n in N*, I don't see how I can arrive at a contradiction.

Answer 1

H is invalid in this frame. Let's take p as false throughout N*, and p as true throughout N. That makes H false in any world in N*. Instead of proving H to be valid in the frame, which it isn't, the correct method is to prove that every substitution instance of H is valid in the model based on this frame in which V(p,0*)=1 for every propositional variable. Then because every substitution instance of H is valid in the model, every theorem of H is valid in the model.