Let me restate the theorem for variables.

**Notation** Read $p\bowtie q$ as "$p$ is adjacent to $q$", and $p\not\bowtie q$ 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 $v_1\neq v_2 \land v_1\not\bowtie v_2$ and $\deg v_1 + \deg v_2 \ge n\in\mathbb{N}$. 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 $p_k\bowtie p_1$ then $p_{k-1} \not\bowtie p_n$, 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.