(Modal Logic) Show that a binary relation is Euclidean if and only if it is transitive and symmetric.

Question

I am attempting that a binary relation is Euclidean if and only if it is transitive and symmetric in a Modal setting, where a modal system is Euclidean if it contains Axiom 5: $\diamond \phi \rightarrow \Box \diamond \phi$ is Euclidean, a modal system is symmetric if it contains Axiom B: $\phi \rightarrow \Box \diamond \phi$, and a modal system is transitive if it contains Axiom 4: $\Box \phi \rightarrow \Box \Box \phi$.

I've made a bit of headway in proving the forward direction: if the modal system is Euclidean, then if $wRv \land wRu$, we have $vRu$. Also, if we have $wRu \land wRv$, we have $uRv$. Because $(wRv \land wRu) \iff (wRu \land wRv)$, we have $vRu \iff uRv$, which shows that the accessibility relation between $v$ and $u$ is symmetric. However, I can't figure out how to show that the accessibility relations between $w$ and both $u$ and $v$ are symmetric, and I can't make any progress on showing transitivity.

Answer 1

You're having trouble because it's not true: not every Euclidean relation is symmetric and transitive. What is true is that a symmetric relation is Euclidean if and only if it is transitive. Is it possible you wanted to show that a relation is symmetric and Euclidean if and only if it is symmetric and transitive?

Aside from this equivalence, there are no other implications between these three properties. It's a good exercise to find relations which are:

• Euclidean and transitive but not symmetric
• Symmetric but neither Euclidean nor transitive
• Euclidean but neither transitive nor symmetric
• Transitive but neither Euclidean nor symmetric