Temporal Logic Tautology

Question

I have the following question:

if it is necessary that p -> p = tautology?

I think it's not, and I am showing my example for the contradiction, below:

And I'll explain:

For every "universe" which is reachable for S1, p is true (hence םp).

For a universe that isn't reachable for S1, p is false (and therefore, this is not a tautology).

But are my observations correct?

Thanks,

Alan

Answer 1

Your observation is not correct.   $\Box p \to p$ does not say "if $p$ is true in all accessible worlds it is true in all worlds (even the inaccessible ones)"

It just says "if $p$ is true in all worlds accessible from this one, then it is true in this world;" for an given world of interest. Which is true if and only the world of interest is accessible to itself.

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Still the point remains that necessity and possibility are relative concepts.

By definition, a predicate is necessarily true if and only if the predicate is true in all accessible worlds.   This is always evaluated relative to some world of interest.  When the world is omitted, we're usually interested in the "actual world" by default.

So $w \vDash \Box p\to p \;$ says, if predicate $p$ is true in all worlds accessible to world $w$ then $p$ is true in world $w$.

However, this is not a tautology; it is the Axiom of Reflexivity.   It only holds in models where the accessibility relation is reflexive.   (Ie: When every world in the frame can access itself.)