# Counter-examples in modal logic

I’m getting stuck on counter-examples to formulas in modal logic. I’m now doing the exercises in Girle’s Modal logic and Philosophy, 3.4.1. Let’s take (a), since the answer to that one is given.

The assignment is to provide a S5 counter-example for

Necesarily (p or (q and (r or Ls)))

I proceed as follows:

Not Necessarily (p or (q and (r or Necessarily s))), at world_0 Possibly not (p or (q and (r or Ls))), at world_0 Not (p or (q and (r or Ls))), at world_1 Not p, at world_1 Not (q and (r or Necessarily s)), at world_1 (here the tree branches into two options) 6.1 Not q, at world_1 6.2 Not (r or Necessarily s), at world_1 7. Not r, at world_1 8. Not Necessarily s, at world_1 9. Possibly not s, at world_1 10. Not s, at world_2

This tree doesn’t close. The counter-example, as far as I can see, is one where at world_1, v(p) = 0; at world_1, v(q)=0, at world_1, v(r)=0 and at world_2, v(s)=0.

Unfortunately, Girle gives this answer: v(p)=v(q)=v(r)=v(s)=0 ALL at the same world, w_0. How did he establish that all these are false at the same world? I don’t get it.

• You don't need to establish that v(p)=v(q)=v(r)=v(s)=0 all at the same world; you can simply assume it because p,q,r,s are all atomic propositions. You don't have to prove that such a world exists. It's enough to show that such a world is possible. – Eliran Jan 11 at 17:41