Proof. Since $F-W$ is a proper subgraph of $F$ with at least three vertices and $F$ is strictly $2$ -balanced,

\begin{align*} \frac {e_F - 1 - \bar {e}_F(W)}{v_F-2-|W|} = \frac {e_{F-W}-1}{v_{F-W}-2} \lt m_2(F) = \frac {e_F-1}{v_F-2}, \end{align*}

which means that $(e_F-1)\cdot |W|\lt (v_F-2) \cdot \bar{e}_F(W)$ and the result follows.

Proof. Since $\delta (G) \leqslant 2e_G/v_G \leqslant 2k+2{\varepsilon } \lt 2k+2$ , we have $\delta (G) \leqslant 2k+1$ and, if ${\varepsilon } \lt 1/2$ , then $\delta (G) \leqslant 2k$ . We may thus assume that $\delta (G) = 2k+1$ and ${\varepsilon } \geqslant 1/2$ , since otherwise (Da) holds. We may further assume that all neighbours of every vertex of degree $2k+1$ have degrees at least $2k+3$ , since otherwise (Db) holds.

Assign to each $v \in V(G)$ a charge of $\deg (v) - 2(k+{\varepsilon })$ and note that the average charge is non-positive. We define the following discharging rule: every vertex of degree $2k+1$ takes a charge of $\frac{2{\varepsilon }-1}{2k+1}$ from each of its neighbours. By our assumption, no vertex of degree $2k+1$ or $2k+2$ sends charge to any of its neighbours. In particular, the final charge of a vertex of degree $2k+1$ is

\begin{align*} 2k+1-2(k+{\varepsilon })+(2k+1) \cdot \frac {2{\varepsilon }-1}{2k+1} = 0 \end{align*}

and the final charge of a vertex of degree $2k+2$ is $2k+2 - 2(k+{\varepsilon }) \gt 0$ .

Since the total charge remains unchanged, the final charge of some vertex of degree at least $2k+3$ must be non-positive. Let $v$ be one such vertex. Suppose that $\deg (v) = 2k+t$ , where $t \geqslant 3$ , and that $v$ has $x$ neighbours with degree $2k+1$ . Since the final charge of $v$ is

\begin{align*} 2k+t - 2(k+{\varepsilon }) - x \cdot \frac {2{\varepsilon } - 1}{2k+1} \leqslant 0, \end{align*}

we have

\begin{align*} (t-2)(2k+1) \lt \frac {t - 2{\varepsilon }}{2{\varepsilon } - 1} \cdot (2k+1) \leqslant x \leqslant 2k+t, \end{align*}

which implies that $t \lt 3 + 1/k \leqslant 4$ . Therefore, $t=3$ and $x \gt 2k+1 \geqslant 3$ , which means that some vertex of degree $2k+3$ has more than three neighbours of degree $2k+1$ . Moreover, we also have $\frac{3-2{\varepsilon }}{2{\varepsilon }-1} \cdot (2k+1) \leqslant 2k+3$ , which implies that ${\varepsilon } \geqslant \frac{4k+3}{4k+4} \geqslant \frac{7}{8}$ .

Proof of Proposition 4.5 in the case $ \mathbf{v}_{\mathbf{{F}}} \boldsymbol{\leqslant} \mathbf{5}$ . Our goal is to prove that no graph $G$ with $e_G / v_G \leqslant m_2(F)$ is an $F$ -core. Suppose by contradiction that $G$ is such a graph and write $m_2(F) = k +{\varepsilon }$ , where $k \geqslant 1$ is an integer and ${\varepsilon } \in [0,1)$ . We split the argument into three cases, depending on which item of Lemma4.7 holds.

Case 1. Suppose that $G$ has a vertex $v$ of degree at most $2k$ and let $e$ be an edge incident to $v$ . By assumption, there are two copies of $F$ in $G$ whose sole intersection is the edge $e$ and thus $\deg (v) \geqslant 2\delta (F) - 1$ . However, since $v_F \gt 3$ , we may apply the Helpful Lemma to learn that $\delta (F) \gt m_2(F) \geqslant k$ , which implies that $\deg (v) \geqslant 2k+1$ , a contradiction.

Case 2. Suppose now that $G$ has an edge $uv$ with $\deg (u) = 2k+1$ and $\deg (v) \leqslant 2k+2$ , and that ${\varepsilon } \geqslant 1/2$ . By our assumption, there are two copies of $F$ in $G$ that intersect only in $uv$ . Since $v_F \gt 4$ , we may use the Helpful Lemma to learn that the endpoints of every edge of $F$ touch strictly more than $2(k+{\varepsilon }) \geqslant 2k+1$ edges. Because the copies of $F$ intersect only in $uv$ , this means that $\{u,v\}$ must touch at least $2(2k+2) - 1 = 4k + 3$ edges in $G$ . However, our assumption implies that $\{u, v\}$ touches at most $\deg (u) + \deg (v) - 1 \leqslant 4k+2$ edges, a contradiction.

Case 3. Finally, suppose that $G$ has a vertex $v$ of degree $2k+3$ connected to two vertices $u_1, u_2$ of degrees $2k+1$ , and that ${\varepsilon } \geqslant \frac{7}{8}$ . We may assume that $u_1 u_2 \notin G$ , as otherwise we are in Case 2. By our assumption, we find two copies of $F$ that intersect only in $u_1v$ . If none of them use $u_2v$ , then we may argue exactly as in Case 2, as the edge $u_1v$ is the sole intersection of two copies of $F$ in the graph $G^{\prime} := G \setminus u_2v$ and $\deg _{G^{\prime}}(u_1) = 2k+1$ and $\deg _{G^{\prime}}(v) =2k+2$ .

We may thus assume that one of the two copies, say $F_1$ , contains $u_2v$ while the other, $F_2$ , does not. Our assumption that $G$ is a core supplies another copy of $F$ , say $F_3$ , that intersects $F_1$ precisely in the edge $u_2v$ . Similarly as before, we will show that there are not enough edges in $G$ that touch $\{v, u_1, u_2\}$ to support these three copies of $F$ . Since $\varepsilon$ , which is the fractional value of $m_2(F)$ , is at least $7/8$ , the graph $F$ must have at least ten vertices, and we are justified in applying the Helpful Lemma to sets $W$ of up to three vertices. In particular, we may conclude that $F_1$ contains at least $\left \lceil 3(k+{\varepsilon }) \right \rceil = 3k + 3$ edges touching $\{v, u_1, u_2\}$ , that $F_2$ contains at least $\left \lceil 2(k+{\varepsilon }) \right \rceil = 2k+2$ edges touching $\{v, u_1\}$ , and that $F_3$ contains at least $\left \lceil k+{\varepsilon } \right \rceil = k+1$ edges touching $u_2$ . Since, among the edges mentioned, only $u_1v$ and $u_2v$ were counted twice, the set $\{v, u_1, u_2\}$ must touch at least $6k+4$ edges of $G$ . However, the bounds on the degrees of $v$ , $u_1$ , and $u_2$ imply that they touch only $\deg (u_1) + \deg (u_2) + \deg (v) - 2 = 6k+3$ edges.

Proof. Suppose that $G$ is a connected $K_4$ -core with $e_G/v_G \leqslant m_2(K_4) = 5/2$ . Since $G$ must satisfy $\delta (G) \geqslant 2\delta (K_4)-1 = 5 \geqslant 2e_G/v_G$ , see Case 1 in the proof of Proposition4.5, $G$ must be $5$ -regular. Now, let $a_1 a_2 \in G$ be an edge, and let $b_1 b_2 a_1 a_2$ and $c_1 c_2 a_1 a_2$ be two copies of $K_4$ whose edge sets intersect solely in $a_1 a_2$ ; in particular, the vertices $b_1, b_2, c_1, c_2$ are all distinct.

Now, the edge $a_1 b_1$ should also participate in two copies of $K_4$ where it is the sole intersection, $a_1 b_1 v v^{\prime}$ and $a_1 b_1 u u^{\prime}$ . However the vertex $a_1$ is of degree $5$ so $\{v,v^{\prime},u,u^{\prime}\} = \{a_2, b_2, c_1, c_2 \}$ meaning that $b_1$ is adjacent to $c_1$ and $c_2$ . Repeating the same argument for $a_1 b_2$ , we get that $b_2$ is adjacent to $c_1$ and $c_2$ meaning that $\{a_1, a_2, b_1, b_2, c_1, c_2\}$ induces a $K_6$ . Since $G$ is connected and $5$ -regular, we have $G = K_6$ .

Proof. Suppose that $G$ is a connected $C_4$ -core with $e_G/v_G \leqslant m_2(C_4) = 3/2$ . Since $\delta (G) \geqslant 2\delta (C_4)-1 = 3 \geqslant 2e_G/v_G$ , the graph $G$ must be $3$ -regular.

Proof. Suppose that a graph $G$ is minimally (list-)Ramsey for $K_3$ and satisfies $e_G/v_G \leqslant$ $m_2(K_3) = 2$ . If, for some $v \in V(G)$ , there was an orientation of the edges of $G[N(v)]$ with maximum out-degree at most one, we could then extend every $K_3$ -free colouring of $G - v$ to $G$ as follows: For every $u \in N(v)$ , the edge $uv$ gets a colour that is different from the colour of the out-edge from $u$ . (Since each triangle involving $v$ contains an edge of $G[N(v)]$ , this colouring is $K_3$ -free.) As every graph with at most four edges has such an orientation, we may assume that $e(N(v)) \geqslant 5$ for every $v \in V(G)$ ; in particular, $\delta (G) \geqslant 4 \geqslant 2e_G/v_G$ , so $G$ must be $4$ -regular.

If $e(N(v)) \gt 5$ for some $v \in V(G)$ , then $G = K_5$ , since $G$ is $4$ -regular and connected. We may thus further assume that $e(N(v)) = 5$ for every $v$ . Pick some $v$ and denote $N(v) = \{u_1, u_2, u_3, u_4\}$ so that $u_1u_3 \notin G[N(v)]$ . Since $G$ is $4$ -regular, there must be a $w \in V(G) \setminus (\{v\} \cup N(v))$ such that $N(u_1) = \{v, w, u_2, u_4\}$ . Moreover, $w$ is not adjacent to either of $v$ , $u_2$ , and $u_4$ , as they all have $4$ neighbours in $\{v\} \cup N(v)$ , and thus $e(N(u_1)) \leqslant 3$ , a contradiction.

Proof. Suppose that this were not true and let $G$ be a smallest graph with maximum degree three that does not admit such a cycle-free colouring for some choice of lists of size two assigned to its edges. If some vertex $v$ had degree at most two or its three incident edges do not all have the same lists, then we could extend any colouring of $G-v$ by assigning different colours to the edges incident with $v$ ; this way we would not create any monochromatic cycles. Therefore, $G$ must be $3$ -regular and, as $G$ is clearly connected, all lists must be identical, say $\{\text{red}, \text{blue}\}$ . Let $v_1 v_2 \dotsc v_\ell v_1$ be an arbitrary cycle in $G$ and denote by $e_1, \dotsc, e_\ell$ the edges incident with $v_1, \dotsc, v_\ell$ , respectively, that are not on this cycle. We can extend any cycle-free colouring of $G-\{v_1, \dotsc, v_\ell \}$ to a cycle-free colouring of $G$ by colouring all the edges of the path $v_1 \dotsc v_\ell$ and the edge $e_1$ red and the edges $v_1 v_\ell$ and $e_2, \dotsc, e_\ell$ blue.

Proof. Given an assignment of lists of size two, pick a random colouring uniformly from the lists. There are $\binom{6}{4} = 15$ copies of $K_4$ in $K_6$ , so the expected number of monochromatic $K_4$ s is at most $15 \cdot 2^{1-6} \lt 1$ , meaning that there is a proper list-colouring.

Proof. If some colour, say red, contains a $5$ -cycle, then we may colour this $5$ -cycle red and the complementary $5$ -cycle not red. If some colour class, say red, contains an edge, say $e$ , not in a triangle, then we may colour $K_5 \setminus e$ without monochromatic triangles (this is possible as no proper subgraph $K_5$ is list-Ramsey for $K_3$ , by Claim4.10) and colour $e$ red. If none of the above is true, then each colour induces one of the following graphs: $K_3$ , $K_4$ , $K_4^-$ , $K_5 \setminus K_3$ , or two triangles sharing a vertex. If some colour, say red, induces $K_5 \setminus K_3$ , then we colour $K_{2,3}$ with red, the remaining edge of $K_5 \setminus K_3$ with not red and the edges of the $K_3$ in the complement with two different colours other than red. If one of the colours, say red, induces $K_4$ or $K_4^-$ , then colour a $C_4$ with red and its diagonal(s) with a colour other than red. Each of the remaining, uncoloured four edges can close at most one monochromatic triangle, as red is not available anywhere outside of the $K_4$ we have already coloured; thus we may colour them one by one. This leaves the case where every colour class is either $K_3$ or two triangles sharing a vertex. But this is impossible, since $3$ does not divide $2e(G) = 20$ .

Proof of Proposition 4.3 . Write $m_2({\mathcal{F}})= k+{\varepsilon }$ , where $k \geqslant 2$ is an integer and ${\varepsilon } \in [0,1)$ , and suppose that $G$ is a connected $\mathcal{F}$ -core with $e_G/v_G \leqslant k+{\varepsilon }$ . It suffices to show that $K_4 \in{\mathcal{F}}$ and $G$ is a $K_4$ -core, since then Claim4.8 will allow us to conclude that $G = K_6$ . We split the argument into three cases, depending on which item of Lemma4.7 holds. Since $K_3$ , $C_4$ , and $K_4$ are the only strictly $2$ -balanced graphs with at most four vertices, the assumption that $m_2({\mathcal{F}}) \gt 2$ implies that the only graph with fewer than five vertices that can belong to $\mathcal{F}$ is $K_4$ .

Case 1. Suppose that $G$ has a vertex $v$ of degree at most $2k$ and let $e$ be an edge incident to $v$ . Since $G$ is an $\mathcal{F}$ -core, there are two copies of some $F, F^{\prime} \in{\mathcal{F}}$ in $G$ that intersect solely in $e$ , and thus $\deg (v) \geqslant \delta (F)+\delta (F^{\prime})-1$ . Since $v_F, v_{F^{\prime}} \gt 3$ , we can apply the Helpful Lemma to learn that $\delta (F), \delta (F^{\prime}) \geqslant k+1$ , which gives $\deg (v) \geqslant 2k+1$ , a contradiction.

Case 2. Suppose now that $G$ has an edge $uv$ with $\deg (u) = 2k+1$ and $\deg (v) \leqslant 2k+2$ and that ${\varepsilon } \geqslant 1/2$ . By assumption, there are two copies of some $F, F^{\prime} \in{\mathcal{F}}$ that intersect only in $uv$ . If $v_F, v_{F^{\prime}} \gt 4$ , we could use the Helpful Lemma to learn that the endpoints of every edge of $F$ and $F^{\prime}$ touch strictly more than $2m_2({\mathcal{F}}) \geqslant 2(k+{\varepsilon }) \geqslant 2k+1$ edges. This would mean that $\{u, v\}$ touch at least $2(2k+2)-1 = 4k+3 \gt \deg (u) + \deg (v) - 1$ edges in $G$ , a contradiction. Thus, one of $F, F^{\prime}$ is a $K_4$ . Consequently, we have

\begin{align*} e_G/v_G \leqslant m_2({\mathcal {F}}) \leqslant m_2(K_4) = 5/2. \end{align*}

As Case 1 rules out $\delta (G) \leqslant 4$ , the graph $G$ must be $5$ -regular and $m_2({\mathcal{F}}) = 5/2$ .

Finally, we show that $G$ is a $K_4$ -core. If this were not true, then some edge $uv$ of $G$ would be the sole intersection of two copies of some graphs from $\mathcal{F}$ , not both of them $K_4$ . The Helpful Lemma implies that, for every $F \in{\mathcal{F}}$ with $v_{F} \gt 4$ , the endpoints of every edge of $F$ touch strictly more than $5$ edges. On the other hand, the endpoints of every edge of $K_4$ touch precisely $5$ edges. This means that at least $10$ edges of $G$ would have to touch $\{u, v\}$ , a contradiction.

Case 3. Finally, suppose that ${\varepsilon } \geqslant 7/8$ and that there are vertices $u, v_1, v_2$ with $\deg (u) = 2k+3$ , $\deg (v_1) = \deg (v_2) = 2k+1$ , and $uv_1, uv_2 \in G$ . We may assume that $v_1v_2 \notin G$ , since otherwise we would be in Case 2. By assumption, there are two copies of some $F_1, F_2 \in{\mathcal{F}}$ that intersect only in $uv_1$ . Applying the Helpful Lemma, we learn that $\delta (F_1) \gt m_2(F_1)$ and $\delta (F_2) \gt m_2(F_2)$ . If $m_2(F_i) \geqslant k+1$ for some $i \in \{1, 2\}$ , then we would have $\deg (v_1) \geqslant \delta (F_1) + \delta (F_2) - 1 \geqslant 2k+2$ , a contradiction. Therefore,

\begin{align*} k+ 7/8 \leqslant k + {\varepsilon } =m_2({\mathcal {F}}) \leqslant m_2(F_i) \lt k+1 \end{align*}

for each $i \in \{1, 2\}$ , and thus the definition of $m_2(\cdot )$ implies that $v_{F_1}, v_{F_2} \geqslant 10$ .

We can therefore apply the Helpful Lemma and learn that the endpoints of every edge of $F_1$ and $F_2$ touch strictly more than $2m_2({\mathcal{F}}) \geqslant 2(k+{\varepsilon }) \geqslant 2k+1$ edges. Because our copies of $F_1$ and $F_2$ intersect only in $uv$ and $\{u,v\}$ touches exactly $4k+3$ edges of $G$ , one of the copies, say of $F_2$ , contains $uv_2$ . By the Helpful Lemma, this copy of $F_2$ must therefore use strictly more than $3m_2(F_2) \geqslant 3(k+{\varepsilon }) \geqslant 3k+2$ edges touching $\{u, v_1, v_2\}$ .

Finally, since $G$ is an $\mathcal{F}$ -core, there is another copy of some graph $F_3 \in{\mathcal{F}}$ that intersects our copy of $F_2$ solely in $uv_2$ . The Helpful Lemma gives $\delta (F_3) \geqslant \lceil m_2(F_3) \rceil \geqslant k+1$ . The union of our copies of $F_1$ , $F_2$ , and $F_3$ must contain at least $3k+3+2k+2+k+1 - 2 = 6k+4$ edges touching $\{u, v_1, v_2\}$ . However, the bounds on the degrees of $u$ , $v_1$ , and $v_2$ in $G$ imply that they touch only $6k+3$ edges, a contradiction.

Proof. If all lists assigned to the edges of $K_6$ are identical, then we can colour so that one colour class is $K_{3,3}$ and the other $2 \cdot K_3$ ; both these graphs have $2$ -density two. We may thus assume that the lists incident to some vertex $v$ are not all identical. Any $\mathcal{F}$ -free colouring of $K_6-v$ (at least one such colouring exists by Proposition4.3) can now be extended to $K_6$ by colouring the edges incident with $v$ so that no colour is repeated more than twice. Since every graph $F$ with $m_2(F) \geqslant 5/2$ has minimum degree at least three, this colouring is also $\mathcal{F}$ -free.

Proof of Proposition 4.4 . Suppose by contradiction that a graph $G$ is minimally list-Ramsey for some family $\mathcal{F}$ of graphs, each containing a cycle, and satisfies $e_G/v_G \leqslant \min \{m_2({\mathcal{F}}), 2\}$ . Observe first that $\delta (G) \geqslant 3$ and that, for every vertex $v$ of degree three, the lists assigned to the edges incident with $v$ are identical. Indeed, if this were not true, we could extend any $\mathcal{F}$ -free colouring of $G-v$ to $G$ by assigning different colours to all edges incident with $v$ ; such a colouring would also be $\mathcal{F}$ -free as every graph in $\mathcal{F}$ has minimum degree at least two. We may also assume that $m_2({\mathcal{F}}) \geqslant e_G/v_G \gt 3/2$ , as otherwise $G$ is $3$ -regular and Claim4.11 implies that it is not list-Ramsey for $\mathcal{F}$ . In particular, since $m_2(C_4) = 3/2$ , the only graphs with fewer than five vertices that can belong to $\mathcal{F}$ are $K_3$ and $K_4$ .

Case 1. $e_G/v_G \in (3/2, 2)$ . We apply Lemma4.7 to $G$ with $k+{\varepsilon } = e_G/v_G$ . Since we have already ruled out the possibility that $\delta (G) \leqslant 2$ , there are only two subcases.

Case 1a. Suppose first that $G$ has an edge $uv$ with $\deg (u) = 3$ and $\deg (v) \leqslant 4$ . Recall that the three lists assigned to the edges incident with $u$ are the same, say $\{\text{red}, \text{blue}\}$ . Consider an arbitrary $\mathcal{F}$ -free colouring of $G-u$ . Since $v$ has at most three neighbours in $G-u$ , it is incident with at most one blue edge or at most one red edge; without loss of generality, this rare colour is blue. Let $w$ be the other endpoint of the blue edge incident with $v$ . Colour two edges incident with $u$ blue and one edge red so that $uw$ , if it is an edge of $G$ at all, is not blue.

If this colouring was Ramsey for $\mathcal{F}$ , we would see a monochromatic copy $F$ of a graph from $\mathcal{F}$ that contains $u$ . As $F$ has minimum degree two, it cannot be red and, if it is blue, it must also contain both $v$ and $w$ and the other blue neighbour of $u$ (which is distinct from $w$ ). However, since $m_2(F) \gt 3/2$ , it cannot contain a path of length three whose two centre vertices have degree two; indeed $\delta (K_4) \gt 2$ and the Helpful Lemma implies that, for every $F \in{\mathcal{F}}$ with $v_F \gt 4$ and every pair $\{a,b\}$ of vertices of $F$ , we have $\bar{e}_F(\{a,b\}) \gt 3$ .

Case 1b. Suppose now that ${\varepsilon } \geqslant 7/8$ and that $G$ has two edges $vu_1$ and $vu_2$ , where $v$ is of degree $5$ and $u_1, u_2$ are non-adjacent vertices of degree $3$ . Recall that, for both $i \in \{1, 2\}$ , the three lists assigned to the edges incident with $u_i$ are identical. Consider an arbitrary $\mathcal{F}$ -free colouring of $G - \{u_1, u_2\}$ . Since $v$ has degree three in this graph, for each $i \in \{1, 2\}$ , there is a colour $c_i$ belonging to the lists seen at $u_i$ that appears on at most one edge of $G - \{u_1, u_2\}$ incident with $v$ (it may happen that $c_1 = c_2$ ). Now, for each $i \in \{1, 2\}$ , colour two edges incident to $u_i$ with $c_i$ (and the third edge in the other available colour) so that there is no triangle whose all edges are coloured $c_i$ ; this is possible as $v$ had at most one incident edge coloured $c_i$ and $u_1$ and $u_2$ are not adjacent.

If this colouring was Ramsey for $\mathcal{F}$ , we would see a monochromatic copy $F$ of a graph from $\mathcal{F}$ that contains $u_i$ for some $i\in \{1,2\}$ . As $F$ has minimum degree two, it must be coloured $c_i$ and it must contain both $v$ and at least one more neighbour of $v$ connected by an edge coloured $c_i$ . If there is only one such neighbour, then $F$ would contain a path of length three whose two centre vertices have degree two, which is impossible (see Case 1a). Otherwise, it must be that $c_1 = c_2$ , there are exactly two such neighbours, and $\bar{e}_F(\{v, u_1, u_2\}) = 5$ . However, since $m_2(F) \geqslant m_2({\mathcal{F}}) \geqslant 1+{\varepsilon } \geqslant 2 - 1/8$ but $F \neq K_3$ , then either $m_2(F) \geqslant 2$ , in which case $\delta (F) \geqslant 3$ , or $15/8 \leqslant m_2(F) \lt 2$ , in which case $F$ has at least ten vertices. In both cases, the Helpful Lemma implies that $\bar{e}_F(\{v, u_1, u_2\}) \geqslant 6$ , a contradiction.

Case 2. $e_G/v_G = 2$ . Since we have already ruled out the possibility that $\delta (G) \leqslant 2$ , either $\delta (G) = 3$ or $G$ is $4$ -regular. Let $u$ be a vertex of smallest degree in $G$ that minimises $e(N(u))$ and consider an arbitrary $\mathcal{F}$ -free colouring of $G-u$ . We will show that it can be extended to an $\mathcal{F}$ -free colouring of $G$ , unless $G = K_5$ . As every $F \in{\mathcal{F}}$ , other than the triangle, has minimum degree at least three, it is enough to show there is a colouring of the edges incident with $u$ that uses each colour at most twice and has no monochromatic triangles containing $u$ . It is straightforward to check that such a colouring exists when $\deg (u) = 3$ , so we will assume that $G$ is $4$ -regular.

Case 2a. $e(N(u)) \leqslant 4$ . Denote the four neighbours of $u$ by $v_1, \dotsc, v_4$ . If the four lists assigned to $uv_1, \dotsc, uv_4$ are identical, say $\{\text{red}, \text{blue}\}$ , then we can colour as follows: Since $N(u)$ is not a clique, we may assume that $v_1v_2 \notin G$ . If $v_3v_4 \in G$ and it is coloured red, colour $uv_3, uv_4$ blue and $uv_1, uv_2$ red; otherwise, colour $uv_3, uv_4$ red and $uv_1, uv_2$ blue. It is straightforward to check that no monochromatic triangle was created.

If the four lists assigned to $uv_1, \dotsc, uv_4$ are not identical, then some colour, say red, appears in the lists of at most two edges incident with $u$ . If red appears only once, say in the list of $uv_1$ , then we colour $uv_1$ red and the remaining three edges as in the case where $\deg (u) = 3$ . Suppose now that red appears in two lists, say of $uv_1$ and $uv_2$ . Since $N(u)$ induces at most four edges, either $v_3v_4 \notin G$ or one of $v_1, v_2$ is not adjacent to one of $v_3, v_4$ . If $v_3v_4 \notin G$ , then we colour $uv_1$ red, $uv_2$ with a colour $c$ different than red and both $v_3$ and $v_4$ with a colour different than $c$ ; it is easy to check that no colour is used more than twice and that no triangle is monochromatic. Otherwise, we may assume without loss of generality that $v_2 v_4 \notin G$ . Now, colour $uv_1$ red, $uv_2$ with a colour $c$ different than red, $uv_3$ with a colour $c^{\prime} \neq c$ , and $uv_4$ with a colour different than $c^{\prime}$ . Again, it is easy to check that no colour was used more than twice and that no triangle is monochromatic.

Case 2b. $e(N(u)) \geqslant 5$ . Since $G$ is connected, $4$ -regular, and $e(N(v)) \geqslant 5$ for all $v \in V(G)$ , it must be $K_5$ , see Claim4.10. Thus, it is enough to show that $K_5$ is not list-Ramsey for any family $\mathcal{F}$ with $m_2({\mathcal{F}}) \geqslant 2$ . To this end, we first claim that every $2$ -balanced graph with $2$ -density at least $2$ that is contained in $K_5$ contains a triangle. Indeed, if $F$ is strictly $2$ -balanced and $m_2(F) \geqslant 2$ , then $v_F \geqslant 3$ and $e_F \geqslant 2(v_F - 2) + 1 = 2v_F-3$ , but every triangle-free graph $F$ satisfies $e_F \leqslant v_F^2/4$ ; it is easy to check that $2v_F - 3 \gt v_F^2/4$ when $v_F \in \{3, 4, 5\}$ . Consequently, any triangle-free colouring of $K_5$ will automatically be $\mathcal{F}$ -free for every family $\mathcal{F}$ with $m_2({\mathcal{F}}) \geqslant 2$ . Finally, the fact that $K_5$ is $2$ -list-colourable with respect to $K_3$ was proved in Claim4.13.

Proof of Theorem 1.7 For the $1$ -statement, it is enough to prove the result for the usual Ramsey property. Recall that $\mathcal{F}$ contains a star forest $F$ where the number of rays in each star is at most $D$ . Let $k$ count the number of stars in $F$ . When $p \gg n^{-(1 + 1/s)}$ , the random graph will a.a.s. contain $r \cdot (k-1) + 1$ disjoint stars with $s$ rays (or indeed, any constant number of copies of this star). By the pigeonhole principle, any $r$ -colouring of the edges of any of these stars will give a monochromatic $D$ -star. By the pigeonhole principle again, at least $k$ of them must use the same colour, giving a monochromatic copy of $F$ .

For the $0$ -statement, assume $p \ll n^{-(1 + 1/s)}$ . At this density, the connected components of $G_{n,p}$ are a.a.s. all trees with at most $s-1$ edges. We claim that one can colour the edges of any graph with this property from any assignment of lists of $r$ colours such that the monochromatic components are all star forests, and the maximum degree in each colour is strictly less than $D$ . This would imply that the colouring does not contain any monochromatic copy of a member of $\mathcal{F}$ .

For every component, we start by taking a vertex $u$ of maximum degree. Colour the edges touching this vertex while making sure that we do not use the same colour $D$ times. Since its degree is at most $s-1 = r\cdot (D-1)$ , we can always do this. Next, if we arrive at a new vertex $v$ , we will colour the new edges that touch $v$ while making sure never to use the colour of the edge that introduced this vertex. This would ensure that all monochromatic components are star forests. In the worst case, this restriction gives us $r-1$ available colours for each new edge.

The same argument would mean that, as long as the number of new edges introduced at $v$ is at most $(r-1)\cdot (D-1)$ , we can still avoid using the same colour $D$ times. Luckily, this is always the case. Indeed, if $v$ introduced at least $(r-1)(D-1) + 1$ new edges, then $\deg (v) \geqslant (r-1)(D-1) + 2$ (counting the edge that led us to this vertex). Recall that by maximality $\deg (u) \geqslant \deg (v)$ . This means that our connected component contains at least $2(r-1)(D-1) + 4 - 1$ edges, and that is strictly greater than $r(D-1) = s-1$ for all $r \geqslant 2$ , giving a contradiction.

Proof. Fix an arbitrary orientation of $G$ with out-degree at most one. If an edge $uv$ , oriented from $u$ to $v$ , is coloured differently than the colour of the out-edge from $v$ (or if $v$ has out-degree zero), we say that $uv$ is nonrepetitive. Our aim is to find a nonrepetitive colouring for all of $G$ , i.e., choose colours so that all edges are nonrepetitive. Since each connected graph that is not a star contains either a path of length three or a triangle and since each orientation of the path of length three and of $K_3$ with out-degree at most one contains a directed path of length two, the colour classes of a nonrepetitive colouring are guaranteed to be star forests.