For x>0, ln(3−2e −x) is clearly positive, monotonic, and bounded from above by ln3, as required by Definition 1. Let Q={a,b,u,v} and let T be the true species tree. To prove that Eq. (1) simplifies to (5), consider two cases. If anchors u,v are not sisters in the species tree quartet on Q (i.e., \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$ab.uv\notin \mathcal {Q}^{T}$\end{document}ab.uv∉QT), by the MSC model, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$p(ab.uv)=\frac {1}{3}e^{-\tau (Q)}$\end{document}p(ab.uv)=13e−τ(Q) and thus, τ(Q)=− ln3p(a b.u v). In the first case in (1), \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$D^{\prime }_{uv}=\beta + \alpha. \tau (Q) = \ln 3+ \tau (Q)= \ln 3 - \ln 3p(ab.uv)=- \ln p(ab.uv)$\end{document}Duv′=β+α.τ(Q)=ln3+τ(Q)=ln3−ln3p(ab.uv)=−lnp(ab.uv). In the second case, u,v are sisters (i.e., \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$ab.uv\in \mathcal {Q}^{T}$\end{document}ab.uv∈QT), and by the MSC model, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$p(au.bv)=1-\frac {2}{3}e^{-\tau (Q)}$\end{document}p(au.bv)=1−23e−τ(Q); thus, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\tau (Q)=-\ln \frac {3}{2}(1-p(au.bv))$\end{document}τ(Q)=−ln32(1−p(au.bv)). In the second case in (1), the distance is β−f(τ(Q))= ln3− ln(3−2e −τ(Q))=− lnp(a b.u v). Thus, in both cases, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$D^{\prime }_{uv} = D^{*}_{uv}$\end{document}Duv′=Duv∗. □