Recall that a logic is a set of wff’s containing all tautologies and closed under modus ponens. Given a logic , a set of wff’s is -consistent if can not be deduced from given . itself is said to be -consistent if can not be deduced from the empty set. Let be a consistent normal modal logic. The canonical frame for is the Kripke frame , where
is the set of all maximally consistent sets, and
iff implies for any wff .
If we define , then the second condition above reads iff .
The canonical model of based on is the pair , where
The main result regarding the canonical model of is:
iff , where is any wff.
which is the result of the following:
For any world in , iff for all worlds such that .
Suppose and . Then by the definition of . Conversely, suppose for all such that . In other words, for all such that , or . But , the deductive closure of , so , and therefore (see here (http://planetmath.org/SyntacticPropertiesOfANormalModalLogic)), or (since ), or (since is maximally consistent). ∎
Proof of Proposition 1. We do induction on the number of logical connectives in . If , then is either a propositional variable or . The former is just the definition of . The later case is just the definition of -consistency. Next, if is , then iff either or iff or iff or iff iff . Finally, if is , then iff iff for all such that iff for all such that .
Recall that a logic is complete in a frame if it is complete in every model based on the frame. As a corollary to Theorem 1, we have
is complete in its canonical frame .
Any wff valid in every model based on is valid in in particular, and therefore a theorem of by Theorem 1. ∎
The converse is not true. There are in fact normal modal logics that are sound in no frames at all.
Canonical models are useful in proving the completeness theorems for many common normal modal logics. To prove that a logic is complete in a class of frames, by the corollary above, it is enough to show that the canonical frame is in the class. Here are two examples:
Let be the smallest normal logic containing the schema . Then is complete in the class of null frames.
Again, we show that is weak identity. Suppose . Then for any , implies that . Now, if , then applying modus ponens to , we get that since is closed under modus ponens. But this means that . So . But since both and are maximal, they are the same. ∎
|Date of creation||2013-03-22 19:35:01|
|Last modified on||2013-03-22 19:35:01|
|Last modified by||CWoo (3771)|