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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

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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 $F=<W,R>$ with $F \models \diamond \square A \rightarrow A$, it must hold that forall models of the frame $M=<W,R,V>$, $M \models \diamond \square A \rightarrow A$

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