# Modal logic and validity (Hughes and Cresswell)

I'm trying to do question 16.3 on page 310 of Hughes and Cresswell for any of you that have access to the text. (This is not for a class, just for my own personal studies.)

16.3: Using the semantics discussed on p291 in which $V_\mu^*(\forall x\alpha, w)=1$ provided $V_\rho(\alpha,w)=1$ for every x-alternative of $\mu$ for which $V_\rho^*(\alpha,w)$ is defined, show that MP is not validity-preserving.

If I'm understanding correctly this is the semantics that doesn't validate $\forall x(\phi x\wedge L\phi x)\to\phi y$ because it is false in a model with $wRw'$ and y is assigned to some $u\in D_w$ but $u\notin D_{w'}$ supposing u does not satisfy $\phi$ in w, but every other individual satisfied $\phi$ in all worlds. $\phi y$ is false at w. But the antecedent is true because although the conjunction is undefined in w it is true where defined. (It is undefined at w because $L\phi x$ is undefined at w when x is assigned u since $\phi x$ (as so assigned) is undefined at $w'$.)

I think I'm misunderstanding something here---perhaps just what is being referred to on page 291---because it seems like assuming $\vdash\alpha\to\beta, \alpha$ and $\not\vdash\beta$ we get a contradiction. If $\beta$ isn't valid, it is false in some world of some model of a frame of system S. Either $\alpha$ is defined at that world or it is not. If it is, then it is true at that world because it is valid. But then the conditional $\alpha\to\beta$ is false at the world, which is in contradiction with our assumption that it is valid. If it is not defined, then it is still true at all worlds in which it is defined and so true at our world (I think this is part of the semantics being referred to), again showing that the conditional is false contradicting our assumption that it is valid.

What am I messing up?