< {\mathtt{w}}'(z)$ and ${\mathtt{w}}''(v) = {\mathtt{w}}'(v)$. Similarly, if ${\mathtt{w}}'(z) + {\mathtt{w}}'(v) < {\mathtt{ub}}'$, then ${\mathtt{w}}''(z) + {\mathtt{w}}''(v) < {\mathtt{ub}}'$. Assume that ${\mathtt{w}}'(z) + {\mathtt{w}}'(v) > {\mathtt{lb}}'$. If $v$ is a neighbor of $z$, then ${\mathtt{w}}'(v) \ge \alpha$ holds, and thus ${\mathtt{w}}''(z) + {\mathtt{w}}''(v) > {\mathtt{lb}}'$ as follows: $$\begin{aligned}
  {\mathtt{w}}''(z) + {\mathtt{w}}''(v)
  &= {\mathtt{w}}'(z) + {\mathtt{w}}'(v) - \varepsilon/2 
  \\
  &\ge {\mathtt{w}}'(z) + {\mathtt{w}}'(v) - ({\mathtt{w}}'(z) + \alpha - {\mathtt{lb}}')/2 
  \\
  &\ge {\mathtt{w}}'(z) + {\mathtt{w}}'(v) - ({\mathtt{w}}'(z) + {\mathtt{w}}'(v) - {\mathtt{lb}}')/2 
  \\
  &= ({\mathtt{w}}'(z) + {\mathtt{w}}'(v) + {\mathtt{lb}}')/2
  \\
  &> ({\mathtt{lb}}' + {\mathtt{lb}}')/2
  \\
  &= {\mathtt{lb}}'.\end{aligned}$$ If $v = z$, then ${\mathtt{w}}'(z) = {\mathtt{w}}'(v) > {\mathtt{lb}}'/2$, and thus $$\begin{aligned}
  {\mathtt{w}}''(z) + {\mathtt{w}}''(v)
  &= 2{\mathtt{w}}'(z) - \varepsilon
  \\
  &\ge 2{\mathtt{w}}'(z) - |{\mathtt{w}}'(z) - {\mathtt{lb}}'/2|
  = 2{\mathtt{w}}'(z) - ({\mathtt{w}}'(z) - {\mathtt{lb}}'/2)
  \\
  &= {\mathtt{w}}'(z) + {\mathtt{lb}}'/2
  \\
  &> {\mathtt{lb}}'.\end{aligned}$$ Next, assume that ${\mathtt{w}}'(z) + {\mathtt{w}}'(v) > {\mathtt{ub}}'$. If $v$ is a non-neighbor of $z$, then ${\mathtt{w}}'(v) \ge \beta$. Thus, ${\mathtt{w}}''(z) + {\mathtt{w}}''(v) > {\mathtt{ub}}'$ holds as follows: $$\begin{aligned}
  {\mathtt{w}}''(z) + {\mathtt{w}}''(v)
  &= 
  {\mathtt{w}}'(z) + {\mathtt{w}}'(v) - \varepsilon/2 
  \\
  &\ge {\mathtt{w}}'(z) + {\mathtt{w}}'(v) - ({\mathtt{w}}'(z) + \beta - {\mathtt{ub}}')/2 
  \\
  &\ge {\mathtt{w}}'(z) + {\mathtt{w}}'(v) - ({\mathtt{w}}'(z) + {\mathtt{w}}'(v) - {\mathtt{ub}}')/2 
  \\
  &= ({\mathtt{w}}'(z) + {\mathtt{w}}'(v) 