Let $K_n$ be a complete graph of $n$ vertices and there exists two complete sub graphs, namely, $K_i\subseteq K_n$ and $K_j\subseteq K_n$ such that $K_n=K_i\cup K_j$. Then either $K_n=K_i$ or, $K_n=K_j$. Is it always true? Justify.
Here, the graphs are considered to be simple and undirected such that the union of two complete graphs $K_i$ and $K_j$ are defined as: $K_i\cup K_j=\langle V(K_i)\cup V(K_j), E(K_i)\cup E(K_j)\rangle$. As many counter examples as i considered so far seem to satisfy the above statement. Can anyone help me find a counter example in which neither $K_n=K_i$ nor $K_n=K_j$, satisfying the above graphs union?
Suppose neither $K_i$ nor $K_j$ are trivial. Then, there has to exist an edge between a node in $K_i$ to one in $K_j$. Is that possible?