Show that the ramsey number $r(K_4,K_5) \leq 25$.
EDIT: The original problem wording is as follows, Show that if $G$ is a graph of order 25 that does not contain $K_4$ as a subgraph then $G$ contains 5 vertices no two of which are connected.
My attempt: Firstly notice that $$deg_{G}(u)+deg_{\overline{G}}(u)=24$$ Therefore by the pigeonhole principle we obtain that either, $deg_{G} \geq 12$ OR we have $deg_{\overline{G}}(u) \geq 12$. Now WLOG we consider the case when $deg_{G}(u) \geq 12$. Denote $N(u)=\{v_1,v_2 \cdots v_{12}\}$ to be the set of vertices of that are adjacent to the vertex $u$. Now we notice that between the 12 vertices if we get a $K_3$ then including $u$ we get a $K_4$ so we consider the ramsey number $r(3,5)=14$. However we only have 13 vertices so we cannot conlude anything however this does rule out all cases in which $deg_{G}(u) \geq 13$ and so we conclude that $deg_{G}(u)=deg_{\overline{G}}(u)=12$ for all $u \in V(G)$. Now from here is where i'm unsure how to contine, any help is appreciated!