This problem is similar to Exercise 8. The difference is that in that exercise we considered a particular structure $\mathfrak{A}$ , while here we specify a criterion for $\mathfrak{A}$ to be a model of some set of sentences $\Sigma$ .

Suppose such $\Sigma$ exists. For each $n$ , consider a sentence $\tau_{n}$ which is $$\begin{eqnarray*} \forall v_{0}\ldots\forall v_{n} & & v_{0}=c\rightarrow(\neg v_{0}=v_{1}\rightarrow Pv_{0}v_{1})\rightarrow\ldots\\ & & \ldots\rightarrow(\neg v_{n-1}=v_{n}\rightarrow Pv_{n-1}v_{n})\rightarrow\neg v_{n}=d. \end{eqnarray*}$$ $\tau_{n}$ says that there is no path from $c^{\mathfrak{A}}$ to $d^{\mathfrak{A}}$ of length $n$ or less. Then, as in Exercise 8, $\mathfrak{A}_{ij}$ is a model of $\tau_{n}$ iff $|i-j|>n$ . Moreover, in $\mathfrak{A}_{ij}$ , $<c^{\mathfrak{A}_{ij}},d^{\mathfrak{A}_{ij}}>$ belongs to the transitive closure of the relation $P^{\mathfrak{A}_{ij}}$ . Therefore, $\mathfrak{A}_{ij}$ is a model of $\Sigma$ . This implies, that there is a model for every finite subset of $\overline{\Sigma}=\Sigma\cup\{\tau_{n}\}_{n\in\mathbb{N}}$ , and, hence, by the compactness theorem, there is a model $\mathfrak{B}$ of $\overline{\Sigma}$ , and, hence, $\Sigma$ . But in this model there is no finite path from $c^{\mathfrak{B}}$ to $d^{\mathfrak{B}}$ , i.e. $<c^{\mathfrak{B}},d^{\mathfrak{B}}>$ does not belong to the transitive closure of the relation $P^{\mathfrak{B}}$ . The contradiction shows that there is no such $\Sigma$ .