First order logic - each vertex of graph lies on some cycle We consider only undirected graphs. I would like to show that there exists sentence $\phi$ in FOL such that $G\models \phi$ if only and only each vertex of $G$ lies on some cycle. Problem for me is that this graph ma be infinite. When it comes to finite case it is easy to give $\phi$: $$\forall_{u \in V}\deg(u)\ge 2$$, what may be expressed in FOL. It is easy to show that every vertex (in finite graph) lies on some cycle only and only each vertex of the graph has degree at least $2$. However, I can't deal with infinite graph. In particularity, $\phi$ must be true both finite and infinite graphs. Of course if in infinite graph each vertex lies on some cycle then we can say that each vertex has degree at least $2$. I can't prove that If in infinite graph each vertex has degree at least $2$ t then each vertex lies on some cycle. Can you help me ?

This isn't true, by a compactness argument.

Suppose that $\varphi$ be a sentence of first-order logic such that for every finite graph $G,$ $G\models\varphi$ iff every vertex of $G$ lies on some cycle. We'll show that this can't be true also for every infinite graph.

Expand your language by adding a new constant symbol $c,$ and consider the theory $T=\lbrace \varphi \rbrace \cup \lbrace "\\!c\text{ does not lie on a cycle of length exactly }n\\!" \mid 0\lt n\lt \omega\rbrace.$

Then every finite subset of $T$ is satisfiable (by, for example, a finite graph consisting of one large cycle of length longer than any of the lengths mentioned in the particular finite subset of $T).$

Therefore $T$ is satisfiable. But any model of $T$ is a graph which satisfies $\varphi$ but in which the interpretation of the constant symbol $c$ lies on no cycle.