Modal Logic: Why are Universal frames a subset of Equivalence frames?


I'm looking through the lecture notes for my course on modal logic and am having a hard time understanding why it is that U, the class of all Universal frames, is a subset of E, the class of all Equivalence frames. I understand that all universal relations are equivalence relations and that not all equivalence relations are universal, but then do not see how the following assertion follows:

"If φ is valid in E, then since U is a subset of E, φ is valid in U." Wouldn't it be possible that φ is valid in U but not in E, since not there are some relations in E that are not in U?

I guess I'm thinking of U being a subset of E as the implication, "If U then E," which perhaps is not the correct way of thinking about it? Alternatively, I suppose we can consider the set of binary relations in each frame; then, if for x to be a subset of y is for all of the members of x to be members of y, wouldn't we have that E is a subset of U? As all of the binary relations in E are in U but U has many more relations?

Thanks in advance for any insight into the question itself and into the definitions involved.


Posted 2019-09-20T21:39:09.463

Reputation: 49



A sentence p is valid in a class of frames C iff p is true in every frame in the class. The point, informally, is that when there are more frames there are fewer sentences true in all of them. In particular, if C is a subclass of D then every sentence true in every frame in D is also true in every frame in C (since every frame in C is a frame in D), so anything valid in D is valid in C, but there might be sentences not valid in D which only fail in frames in D-C so are valid in C.

For a simple example, let C be the class of all reflexive frames, and D the class of all frames. Then C validates the sentence p->Diamond(p), but D doesn't (consider a frame with no arrows at all).

So since U is a subclass of E, the set of sentences valid in U is a superset of the set of sentences valid in E.

Noah Schweber

Posted 2019-09-20T21:39:09.463

Reputation: 2 210