graph theory - Proof of Bondy and Chvátal Theorem

Let me restate the theorem for variables.

Notation Read $pbowtie q$ as "$p$ is adjacent to $q$", and $pnotbowtie q$ as "$p$ is not adjacent to $q$".

Theorem Let $G$ be a graph whose order is greater than $2$. Let $v_1$ and $v_2$ be vertices such that $v_1neq v_2 land v_1notbowtie v_2$ and $deg v_1 + deg v_2 ge ninmathbb{N}$. Then $G$ is Hamiltonian iff $G' = G + v_1 v_2$ is Hamiltonian.

Proof Assume $G'$ is Hamiltonian but not $G$. Therefore, by definition, there exists a cycle $(p_1,ldots,p_n)$ in $G$ connecting $v_1$ (at $p_1$) to $v_2$ (at $p_n$) visiting all of $G$'s vertices. If $p_kbowtie p_1$ then $p_{k-1} notbowtie p_n$, since $(p_1,p_k,p_{k+1},ldots,p_n,p_{k-1},p_{k-2},ldots,p_1)$ is a Hamiltonian cycle of $G$. As such, $deg p_n le n - (1 + deg p_1)$, a contradiction.

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