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Relation $R\subseteq A\times A$ has Rosser-Churchil propetry if for each $a,b,c$ such that there exists path between $a$ and $b$, and from $a$ to $c$ then also exists $d$ reachable from $b$ and $c$. Prove that there exists formula in monadic second order logic that expresses graph with this property

It is fairly easy:
$$\forall_{a}[\{\exists_b\exists_c ((b\neq c) \wedge Path(a,b) \wedge Path(a,c))\} \to (\exists_d (Path(b,d)\wedge Path (c,d))]$$
It should be ok, but the problem is that I can't express $Path$.
In finite graph I can do it using linear relation, but in in this case it seems to be not working. Moreover, I don't know what I should think about infinite path between some $x$ and $y$ - are them reachable each other ?

The second thing is: Maybe, my way is wrong ? I should only prove that such formula exists, no write. Maybe some regular language is equivalent to this property of relation ?
Keep in mind that formula should be of form:
$\text{quantify over sets} \text{*here only first order quantifiers*}$
So, as you can see, quantification over set can happened only at front of formula

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    Well so far you've only used first order logic. So you should try to quatify over predicates to express $Path$. I'm not exactly sure how to express $Path$. But I think you could replace it with $SimplePath(x,y)$ defined as "There is a simple path from $x$ to $y$", where a simple path is a path with no repeating vertex. $SimplePath$ is easier to express. – 2017-02-13
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    Hmm, After all $SimplePath$ is sufficent for us. Try to show it, please – 2017-02-13
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    I know how to show it. But just giving you the answer would be pointless. Try to define $SimplePath(x,y) := \exists P, ...$ where in the $...$ you says that $P$ is a simple path from $x$ to $y$. If you look at it from an undirected point of view, being a path means that every vertex has exactly two neighbours (in the path) except the two endpoints that only have one. Now you can try to build the formula for $SimplePath$ that works for the directed case. – 2017-02-13
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    (Remark: The simple approach that I described above (saying that every node has two neighbours except the two endpoints) allows $P$ to be the union of a simple path from $x$ to $y$ and of several disjoint cycles. But that's not really a problem. This could be fixed by saying that $P$ is the smallest such set, but it's not necessary since you just want to know if a simple path exists) – 2017-02-13
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    Maybe, the best idea is to express: *$Path(x,y)$ iff there exists subgraph such that it contains $x$ and $y$, subgraph is connected (one component*. This is equivalent to existence of path. What dio you think ? – 2017-02-13
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    How would you express being connected without using the notion of Path? It looks hard to me. The nice thing about paths is that you can just check **locally** that it's a path and it suffices to fully characterise a path. A connected set looks way harder to characterise locally. – 2017-02-13
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    Look here, http://math.stackexchange.com/questions/2141454/monadic-second-order-each-component-has-node-satisfying-predicate-p/2142241?noredirect=1#comment4406340_2142241 Author of answer managed to express existence of Path in MSO. Is it similar to your idea ? – 2017-02-13
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    Right. So you have a solution to your problem? And I initially missed that in your question, but as they say in the answer you linked, you only ever talk about finite paths, even in infinite graphs. – 2017-02-13
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    Hmm, @bof's solution doesnt define path, we need in this problem Path. I still don't how to solve uit – 2017-02-13
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53553/discussion-between-xavierm02-and-haskell-fun). – 2017-02-13

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