I'm currently working on exercises of the book "Modal Logic" by A.Chagrov and M.Zakharyaschev (for pleasure, not homework).

One exercise asks to prove this version of Gabbay rule (exercise $3.10$): A frame $F$ validates the rule $(\Box p \rightarrow p) \vee \psi$ $/$ $\psi$, with $p$ not appearing in $\psi$, if and only if $F$ is irreflexive.

(Note that: "$p$ not appearing in $\psi$").

I have the $\leftarrow$ part, but I'm having a hard time proving the other implication. I would really like to understand this excercise, and I know that the excercises in this book are sometimes hard. But I think this has to be fairly easy.

Until now I tried proving it by reductio ad absurdum: I have one reflexive node $x$, and I try to find a formula $\phi$ that $x$ doesn't validate, but that every other node does validate. If I can prove such formula exists then I'm done, because I plug it in the rule.

I hope I was clear, thanks in advance!


I don't have the book, but it looks like a formula $ ConditionQ \to ( (B \lor A)) \to A) $ you are trying to find what $ConditionQ$ is

$ConditionQ$ is just that $ ( (B \lor A)) \to A) $ is a theorem, this means that $B $ has to be false, in this case B is the T axiom (valid in all reflexive frames) so the condition Q is that all frames are irreflexive.


The formula you're looking for is $(\Box p \wedge p)$.

Given formulas $\Box p$, $p$, and $A$, there are 8 ways to construct a world, but only 6 of these worlds are reflexive ($R?$).

[]p   p      A      R?
---   ---    ---    ---
F     F      F      T
F     F      T      T
F     T      F      T
F     T      T      T
T     F      F      F
T     F      T      F   <- world x
T     T      F      T
T     T      T      T

The only way for $(\Box p \rightarrow p) \vee \psi / \psi$ to be a valid inference rule is for $A$ to be true and $(\Box p \rightarrow p)$ to be false. You can see from this table that this is only true for the row indicated world x.

For a world to be reflexive, it must be accessible from itself. Accessibility is determined primarily by the definition of $\Box$. The formula $\Box p(w_0)$ indicates that $p$ is necessarily true in all possible worlds that are accessible from world $w_0$, but it is not necessary for $p$ to be true in $w_0$ itself.

If we want to ensure that every world is accessible from itself, we need to impose the restriction that if $\Box p(w)$ is true, then $p(w)$ must also be true.

That is precisely what the logical connective $p \rightarrow q$ ("p implies q") means. The truth table looks like this:

p     q      p -> q
---   ---    ------
F     F      T
F     T      T
T     F      F
T     T      T

Therefore, to ensure reflexivity, we add the axiom $\forall w.\Box p(w) \rightarrow p(w)$, or simply $\Box p \rightarrow p$, and that is why this is called the reflexive axiom.

  • $\begingroup$ If I understand you correctly, you are missing the "$p$ not appearing in $\psi$" part. $\endgroup$ – Luis N Scoccola Feb 24 '14 at 12:18
  • $\begingroup$ Well, it's implied by the first truth table. For example, if $A = p$, then world x contains a contradiction, because $p$ is true and $A$ is false, so that line would have to be removed from the list (along with all the other lines where $A$ and $p$ don't match). The only way you get all 8 possible combinations of 3 boolean variables is to make the variables completely independent. $\endgroup$ – tangentstorm Feb 24 '14 at 13:23
  • $\begingroup$ I don't get how this solves the problem. You say that the formula I'm looking for is $(\Box p \wedge p)$ but $p$ does appear in that formula. And I can't substitue it for, say, $q$ because then we don't get the desired result. $\endgroup$ – Luis N Scoccola Feb 24 '14 at 23:17
  • $\begingroup$ Perhaps I was wrong about $(\Box p \wedge p)$: I don't really understand the nature of the proof you're trying to write. I don't know what "the $\leftarrow $ part" means or "the other part". All I know is that the only possible way for the statement $(\Box p\rightarrow p) \vee A / A$ to be valid is if $A$ is a theorem and $(\Box p\rightarrow p)$ is an antitheorem. I then showed (by the first truth table) that this is only possible in one situation (world x), and that a frame containing world x cannot be reflexive because $(\Box p\rightarrow p)$ is a necessary condition for reflexivity. $\endgroup$ – tangentstorm Feb 25 '14 at 8:55

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