[FOM] A minor issue in modal logic
rgheck at brown.edu
Sun Jul 4 17:07:25 EDT 2010
A minor correction to my previous post, which follows: The model should
of course have included a relation R of accessibility between worlds. It
doesn't enter into the discussion, however.
On 07/04/2010 12:27 PM, Richard Heck wrote:
> On 07/03/2010 01:53 PM, Keith Brian Johnson wrote:
>> All and sundry: I have encountered an issue in modal logic that I haven't seen and don't know where to look for a resolution of, and my local philosophy department's members haven't been much help. Perhaps FOMers can help.
>> The fundamental question, I think--I'll give its origin in a moment--is this: For any proposition p, where "Np" means "It is necessarily true that p" and "Np[a]" means "p is necessarily true in the actual world," are Np and Np[a] equivalent--or, if not, does Np[a] at least entail Np?
> The semantics of such operators is well understood. The answer, in both
> cases, is "No".
> Let's use "Ap" to mean: p is true in the actual world. The operator "A"
> is known as the "actuality operator". Let's define it carefully. A model
> is a triple<W,V,@>, where W is a set of worlds, V is a map from worlds
> to sentence-letters (true in that world), and @ is the actual world of
> the model. Then "A\phi" is true in a world w \in W iff \phi is true in
> @. Note that the parameter w does not appear on the right hand side. So
> it follows immediately that "A\phi" is true in all worlds or false in
> all worlds, as \phi is true or false in @. That is, "A\phi" is always
> necessarily true or necessarily false at _every_ world, which of course
> implies that "A\phi" is equivalent to "NA\phi". Moreover, since \phi
> implies "A\phi" (in the sense that, if \phi is true in the model, i.e.,
> at @, then "A\phi" is also true in the model), we have that \phi implies
> "NA\phi". So \phi is equivalent to "NA\phi", in the sense that they are
> always true or false in the model (i.e., at @) together. So "NA\phi"
> certainly does not imply "N\phi", since otherwise \phi implies "N\phi".
> So the simplest sort of counter-example is: Take a two world model; let
> every world be accessible to every other. Let p be true at @ but false
> at the other world. Then Ap is true at both worlds (since p is true at
> @), so NAp is true at both worlds, but of course Np is false at both worlds.
Richard G Heck Jr
Romeo Elton Professor of Natural Theology
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