# I want to prove that $L(M(\alpha))$ isn't a theorem in $K$ system

In Cresswell and Hughes: "A new introduction to Modal Logic" they in question 2.4 ask to show that $L(M(\alpha))$ isn't a theorem of $K$ system (where L is the necessary operator and M the possible operator).

Here's what I thought, I can substitute $\alpha$ for p&~p

So I get:

1. L(M(p&~p)) assumption ad absurdum.

2. L(Mp & M~p) 1, K8+N+K+MP

3. LM(p) & LM(~p) 2, K3.

Here's where I got stuck, and I am not sure how to deduce a contradiction.

Any hints or thoughts?

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I see that you are active on math.sx. Could you perhaps give us some context as to why you posted this question here instead of there? (It would seem a good fit for math.sx to me.) –  DBK Nov 23 '12 at 21:13
I think you've spotted a useful trick with using p&~p as a test case, but C/H's proof system doesn't let you use proof by contradiction for its modal component. In fact, it doesn't let you make assumptions at all. Everything needs to be a formal inference from axiomatic starting points. Your best bet is to actually try to prove ~L(M(p&~p)) as a theorem; is it possible to show that L(~p) -> ~L(p), for example?. –  Paul Ross Nov 23 '12 at 21:43
@DBK, I don't think there are experts in modal logic there, but I will try. I know that in CS Theory there's a stackexchange and they have modal logic tag. (though it has only two followers). –  MathematicalPhysicist Nov 24 '12 at 19:16
Why are you distributing your possibilitation operator? Can't you show that L(M(φ)) => M(φ) under $K$? As a rule of thumb, it's advisable to reduce the number of modal operators on a sentence first- M(p & ~p) should be simpler to prove. (I'm perplexed though- what rule or schema are you applying with K8 - and why N? I'm using Chellas's 'old' introduction to ML here, so maybe I'm missing out) –  Ryder Nov 24 '12 at 23:04

Consider a model M, where the set of possible worlds is {w1,w2}, and the accessibility relation R={(w1,w2)}. Then since there are no possible worlds accessible from w2, for any formula A, M(A) is false at w2. Therefore, L(M(A)) is false at w1; hence L(M(A)) is not valid. Given that K is complete, then L(M(A)) is not a theorem for any formula A.