Talk:Hybrid logic

How is @ip any different than $$ i \rightarrow p $$?--SurrealWarrior 23:35, 31 March 2007 (UTC)


 * Consider for example @$$_a (i \rightarrow p)$$ vs @$$_a $$@$$_i p $$. In the first case it is only claimed that p holds in world 'i' if 'a' and 'i' happen to refer to the same world; in the second case there is the claim that p holds in 'i', regardless of what world 'a' refers to. (Note that it is a given each nominal refers to some world). I suppose you could look at it that @$$_i p $$ claims that $$ i \rightarrow p $$ holds everywhere, rather than just at the point of evaluation (i.e. world 'a' in the example).


 * Here is another explanation. These formulae should be evaluated relative to a particular model i.e. a set a states with relations defined over it (although relations do not arise in this example) and a valuation defined on both propositional variables p,q,r and nominals i,j,k. This valuation defines (i) which propositional variables hold in which states and (ii) assigns each nominal to a single state, note a single state may have multiple names.


 * One can evaluate the truth of a hybrid formula relative to a particular state of such a model, also one may talk of global truth if a formula holds in all states of our model. Now if one is talking about global truth, then the two formulae you presented are in fact equivalent. However if you evaluate them at a particular state they differ, as shown in the comment above. In detail, the truth of @ip at a state x tells us that at the state named by i, p holds -- the @-operator allows us to jump to the state named by i. However the truth of $$i \rightarrow p$$ at x means that either (i) x is not named by i or (ii) x is named by i and p holds there. These are clearly rather different, the second meaning does not allow us to jump to other states and perform evaluations there. --Algebran 14 September 2008