# Euclidean Models, Finite Model Property and Filtrations (Modal Logic)

I have partially solved Exercise $$1$$ from this problem set by Yde Venema, and I need some help in completing it.

Blackburn and Venema's Modal Logic 2.3.8: Call a frame or model euclidean if it satisfies $$\forall x\forall y \forall z ((Rxy ∧ Rxz) → Ryz)$$, and let $$E$$ be the class of euclidean models. Fix a formula $$ξ$$, and let $$Σ$$ be the smallest subformula closed set of formulas containing $$ξ$$ that satisfies, for all formulas ψ$$:$$ if $$\diamond ψ ∈ Σ$$, then $$\square\diamond ψ ∈ Σ$$. Note that in general, $$Σ$$ will be infinite.
(a) Prove that $$E \Vdash \diamond ψ → \square\diamond ψ$$.
(b) Prove that every euclidean model can be filtrated through $$Σ$$ to a euclidean model.
(c) Show that every euclidean model satisfies the following modal reduction principles, $$\diamond\diamond\diamond\leftrightarrow \diamond\diamond, \diamond\diamond\square\leftrightarrow \diamond\square, \diamond\square\diamond\leftrightarrow \diamond\diamond$$ and $$\diamond\square\square\leftrightarrow \diamond\square$$.
(d) Prove that the basic modal similarity type has the finite model property with respect to the class of euclidean models.

I was able to prove the first two parts (I need help with c and d), and here's what I did:

(a) Pick an arbitrary $$\mathfrak M \in E$$, and $$w\in W$$ where $$\mathfrak M = (W,R,V)$$. Suppose $$\mathfrak M,w\Vdash \diamond\beta$$. We want $$\mathfrak M,w\Vdash\square \diamond\beta$$. Suppose instead that $$\mathfrak M,w\Vdash\lnot\square \diamond\beta$$, i.e. $$\mathfrak M,w\Vdash\diamond \square\lnot\beta$$. $$\mathfrak M,w\Vdash \diamond\beta$$ iff there is $$v\in W$$ such that $$wRv$$ and $$\mathfrak M,v\Vdash \beta$$. $$\mathfrak M,w\Vdash\diamond \square\lnot\beta$$ iff there exists $$z\in W$$ such that $$\mathfrak M,z\Vdash \square\lnot\beta$$. Now since $$\mathfrak M$$ is Euclidean, we must have $$wRz \land wRv \to vRz\land zRv$$. Now, clearly, $$\mathfrak M,z\Vdash \square\lnot\beta$$ and $$\mathfrak M,v\Vdash \beta$$ cannot hold together. This is a contradiction, and the proof is complete.

(b) This follows from the definition of $$R^f$$ in a filtrated model. We know that $$wRv\implies [w]R^f[v]$$. Suppose $$\mathfrak M$$ is Euclidean. Then, $$\forall x\forall y \forall z ((Rxy ∧ Rxz) → Ryz)$$. Pick arbitrary $$x,y,z$$ such that $$xRy \land xRz$$. Then, $$[x]R^f[y] \land [x]R^f[z]$$ hold. Also, since $$\mathfrak M$$ is Euclidean, we have $$yRz$$ and hence $$[y]R^f[z]$$. Since $$x,y,z$$ were arbitrary to begin with, the filtrated model is also Euclidean, and we are done.

I have trouble proving (c) and (d). The approach to (c) is straightforward though, to show that $$\phi_1\leftrightarrow\phi_2$$ holds, I should assume $$\phi_1$$ at an arbitrary world, prove $$\phi_2$$ and vice versa. However, I am unable to complete the arguments in this case. Could someone help? Also hints for (d) would be good!

Thanks a lot.

For the first part of (c). If $$\diamond \diamond \diamond \phi$$ holds at $$w$$ then we have this picture for $$R$$ $$w \to x \to y \to z$$ where $$\phi$$ holds at $$z$$, and $$\diamond \phi$$ holds at $$y$$, and $$\diamond \diamond \phi$$ at $$x$$
From the Euclidean property we have $$R(w,x) \wedge R(w,x) \to R(x,x),$$ $$R(x,y) \wedge R(x,x) \to R(y,x),$$ $$R(y,x) \wedge R(y,z) \to R(x,z).$$ As $$\phi$$ holds at $$z$$ and $$R(x,z)$$ we get $$\diamond \phi$$ at $$x$$ and so $$\diamond \diamond \phi$$ at $$w$$.
For the converse: Suppose $$\diamond \diamond \phi$$ holds at $$w$$ with $$\phi$$ at $$y$$ in this picture $$w \to x \to y$$ using the above argument that $$R(x,x)$$ we get $$w \to x \to x \to y$$ and thus $$\diamond \diamond \diamond \phi$$ at $$w$$.
For the next part of (c): $$\diamond \diamond \Box \leftrightarrow \diamond \Box$$, suppose $$\diamond \diamond \Box \phi$$ holds at $$x$$. Then there is $$y$$ with $$R(x,y)$$ and $$z$$ with $$R(y,z)$$ where $$\Box \phi$$ holds at $$z$$. Now $$\Box \phi$$ must hold at $$y$$, because if $$R(y,w)$$ then because $$R(y,z)$$ we have $$R(z,w)$$ so that $$\phi$$ holds at $$w$$. Hence $$\diamond \Box \phi$$ holds at $$x$$.
Conversely, suppose that $$\diamond \Box \phi$$ holds at $$x$$ with $$R(x,y)$$ and $$\Box \phi$$ holds at $$y$$. As observed earlier $$R(y,y)$$ must hold, and hence $$\diamond \Box \phi$$ holds at $$y$$. Thus $$\diamond \diamond \Box \phi$$ holds at $$x$$.