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.


1 Answer 1


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 $x$.

I think the rest of part (c) will be rather similar


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