# What is the justification for closing a whole branch in a modal truth tree diagram if there is a contradiction in only one world?

I am watching "Modal logic 1.2 - truth trees for system K" presented through Kane B's channel.

The standard propositional truth tree diagrams are not a problem. Where I am having trouble is opening worlds in the tree diagram and closing branches with these opened worlds. I don't see why I am justified in closing a whole branch if some worlds in that branch do not have a contradiction in them.

At about 14:00 in the video the presenter states, "If we derive a contradiction in either of these worlds the whole branch closes."

Shouldn't we close all worlds prior to closing the whole branch?

Here is a screenshot of the tree:

I don't see why stacking the worlds justifies the step. Creating world "w2" was not permitted without the stacking.

Up to the branch in w0, we have only propositional logic rules.

Then we go on on the right branch first : □¬p.

We apply the modal rule for with ◊p: we need a new world w1 such that w1 ⊩ p.

But from w0 ⊩ □¬p it follows that w1 ⊩ ¬p and we have the contradiction closing the right branch.

Now for the left one with □q; we apply again the modal rule for to ◊¬q.

This step needs a new world w2 such that w2 ⊩ ¬q.

But from w0 ⊩ □q it follows that w2 ⊩ q and we have the contradiction closing also this branch.

"Worlds" behave exactly as the "single universe" of classical logic : they cannot tolerate contradictions.

The basic rule for Kripke semantics for Modal logic is :

v(□A,w)=T iff for every world w′ in W we have : v(A,w′)=T.

The rule states that □A is true (at a world w) exactly when A is true in all possible worlds.

From it, we have the corresponding one for :

v(◊A,w)=T iff for some world w′ in W we have : v(A,w′)=T.

The rule insures that ◊A is true just in case A is true in some possible world.

• +1 The w0 world is propositional logic rules. I think what you are doing with w1 is what he did with w2. I can see how that closes that world (the presenter's w1). Apparently I don't need to even open the first world. I assume all I have to do is find a world with a contradiction in it. – Frank Hubeny Nov 15 '18 at 15:41

A branch represents what would follow if X were the case. If you get contradiction anywhere in that branch, that means that X cannot be the case, so it doesn't matter what happens elsewhere in the same branch.

More specifically, in your example, the branch under consideration represents the possibility that the following are true: `◇p, ◇~q, ◻q`. What follows is that there must be two worlds, `w1` and `w2` (note that it's possible that `w1=w2`, but we can ignore that), such that `w1` contains `p, q`, and w2 contains `~q, q`. But that means that `w2` contains a contradiction, which is impossible. So `◇p, ◇~q, ◻q` together entail a contradiction. Thus the branch representing this possibility must be closed.

• +1 Thank you. I can see that there is a contradiction in w2. Why do I need to open w1? Can I simply open the world, w2, that has a contradiction and close the branch? – Frank Hubeny Nov 15 '18 at 15:33
• @FrankHubeny Sure, you can do that, and that would be more efficient. The presenter probably opened w1 first only for instructional purposes. – Eliran Nov 15 '18 at 15:39