Show that if a self-complementary graph contains a pendant vertex, then it must have at least another pendant vertex.
$\exists v\in V (G):d_G(v)=1 (\because \text{self-complementary graph contains a pendant vertex})$
$\implies \exists w \in V (G):d_G(v)=odd.(\because {\sum_{v\in V (G) }d_G(v)=2.\text{No.of edges}})$ I am not able to prove it has another pendant vertex.