# prove that a frame is symmetric in modal logic

Im trying to prove or refute that ⋄□A→A characterizes symmetry. I can construct a counterexample, where [W,R, V] would be an interpretation W = {w1,w2}, and the accessibility Relations w1 Rw2, w2 Rw3, Hence the Kripke Frame [W,R] is symmetric. Now Im trying to define a variable assignment that would show that symmetry doesnt hold

I think you need to show that it indeed characterizes symmetry.

Since looking at your start formula:

$$\diamond \square A \rightarrow A$$

the contra positive must hold:

$$\neg \diamond \square A \leftarrow \neg A$$

Applying the definitions of $$\square A \leftrightarrow \neg \diamond \neg A$$ and $$\diamond A \leftrightarrow \neg \square \neg A$$ you can derive:

$$\neg \neg \square \neg \neg \diamond \neg A \leftarrow \neg A$$

which then can be simplefied to:

$$\neg A \rightarrow \square \diamond \neg A$$

which looks very similar to the definiton of symmetry in modal logic:

$$P \rightarrow \square \diamond P$$

Also you have a little typo in $$w_2 R w_3$$, which probably should be $$w_2 R w_1$$. I think you can find the rest you need here https://plato.stanford.edu/entries/logic-modal/.

To show that it characterizes symmetry you have to show that $$(F \models \diamond \square A \rightarrow A) \Leftrightarrow (F$$ is symmetric). You can do this by proving each implication direction separately. Remember that for a frame $$F=$$ with $$F \models \diamond \square A \rightarrow A$$, it must hold that forall models of the frame $$M=$$, $$M \models \diamond \square A \rightarrow A$$