0

I'm reading up on .

In Priest's nonstandard $N$ of his "Introduction to Nonclassical Logic [. . .], Second Edition", it is an exercise to

show

$$\not\vdash_N \square p\supset p$$

and exhibit a counterexample in the style of his examples of the logic $N$ ibid.

My Attempt:

Assume $N$. Then the tableau might be

$$\begin{align} \lnot(\square p &\supset p), 0\\ \square p, & 0\\ \lnot p, & 0\\ p, & 0, \end{align}$$

with the diagram for the counterexample being

$$\stackrel{p, \lnot p}{\stackrel{\curvearrowright}{\boxed{w_0}}}.$$

This ought to be a simple exercise for me but, alas, I'm stuck; I think I did it wrong.

Why?

Well, I'm on page 97 ibid and I hadn't done the exercises necessary from the previous chapter, $\S 4$.

I don't have the time to do every exercise of the book. I picked this exercise because it seemed easy.

Please help :)

Edit: It appears that I have shown the negation of the statement in question by mistake. Exactly where did I mess up? Or does the principle of explosion not hold in $N$?

Community
  • 1
Shaun
  • 44,997
  • NB: This is not my only attempt. It's the one I can best communicate here using $\LaTeX$ and the amount of time I have. – Shaun Aug 09 '19 at 23:23
  • NB2: Time permitting, I'll add a description of $N$. I'm only giving the book a cursory look. – Shaun Aug 09 '19 at 23:26
  • 1
    As far as I can tell skimming the beginning of chapter 4, $N$ is weaker than $K.$ There is a standard countermodel showing this is not provable in $K$: a single (normal) node with nothing accessible from it and $p$ false at it. – spaceisdarkgreen Aug 10 '19 at 00:16
  • Thank you, @spaceisdarkgreen. Please would elaborate on that by providing an answer? Perhaps it's the time of night here (01:33 . . . now) or something but I don't follow. – Shaun Aug 10 '19 at 00:33
  • So, are you saying, @spaceisdarkgreen, that the counterexample would be like this: $$\stackrel{\boxed{w_0}}{\lnot p}?$$ – Shaun Aug 12 '19 at 15:24
  • 1
    I'm not super familiar with this notation, but skimming the chapter again it looks like the box indicates a non-normal world. The world should be normal. – spaceisdarkgreen Aug 13 '19 at 00:12
  • Thank you, @spaceisdarkgreen :) – Shaun Aug 13 '19 at 10:18

1 Answers1

3

$\square p\supset p$ is not provable in K: it is an instance of the additional axiom we add to get the stronger system T. Since any normal model is also a non-normal model (i.e. K is stronger than N), the usual countermodel for showing $\square p\supset p$ is not a theorem of K shows that it is not a theorem of N either.

The usual countermodel has a single successor-less (normal) node at which $p$ is false. Since the node is normal and has no successors, $\square p$ is true here, so since $p$ is false, $\square p \to p$ is false.

The first few of these statements in this problem are just statements that are well-known to not be provable in K so really belong in the previous chapter: I think the real purpose of this exercise is as a warm-up the next part where it asks which ones hold if you add reflexivity / transitivity. For instance, we know that in normal modal logic $\square p \supset p$ becomes valid if we add the reflexivity condition... is that still the case if we drop the normality? It is, but we have to be careful to make sure the argument from last chapter generalizes.

(And by contrast, I don't believe the transitivity axiom $\square p\supset \square\square p$ for normal modal logic is valid for transitive non-normal models... don't take my word for that though, I've never seen a presentation of semantics of non-normal modal logic until right now.)

spaceisdarkgreen
  • 58,508
  • The bounty is as good as yours. I'm sorry I doubted your answer. For whatever reason, I'd gotten the impression that (non-)normal worlds hadn't been defined yet. This mistake of mine has me thinking I ought to read the book more carefully. – Shaun Aug 20 '19 at 19:57