We need to show that L<R1 and R1=R2. We remind the reader that all branches are assumed to be strictly positive and that f is a positive and monotonically increasing function bounded from above by β. In all cases, the equality of R1 and R2 is immediately clear from the Table 2. The inequality L<R1 follows directly from the fact that f(x) is monotonically increasing in Cases 1, 2, and 5. For Case 3, because of positivity of f(x) and α, we have L<2β<R. Similarly, for Case 4, L<2β+α e 1<2β+α e 1+2α e 2=R. Case 6 follows from the positivity of f, and Case 7 is trivially correct for positive branch lengths. Thus, we have shown in all possible relationships between {u,v} and the quartet tree, the four point condition holds for the topology consistent with tree T. Therefore, the proof is complete for the double anchored case. Table 2Proof of four-point condition for double anchors. Four point condition for all 7 cases of adding {u,v} to a quartet tree, as shown in Fig. 4 (left side) \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$L =D^{\prime }_{uv}[a,b]+D^{\prime }_{uv}[c,d]$\end{document}L=Duv′[a,b]+Duv′[c,d] \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$ R1 =D^{\prime }_{uv}[a,d] + D^{\prime }_{uv}[b,c]$\end{document}R1=Duv′[a,d]+Duv′[b,c] \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$R2 =D^{\prime }_{uv}[a,c] + D^{\prime }_{uv}[b,d]$\end{document}R2=Duv′[a,c]+Duv′[b,d] Case 1[β−f(e 1+e 3)]+[β−f(e 2+e 3)][β−f(e 3)]+[β−f(e 3)][β−f(e 3)]+[β−f(e 3)]Case 2[β−f(e 1+e 2+e 3)]+[β−f(e 3)][β−f(e 3)]+[β−f(e 1+e 3)][β−f(e 1+e 3)]+[β−f(e 3)]Case 3[β−f(e 1)]+[β−f(e 3)][β+α e 2]+[β+α e 2][β+α e 2]+[β+α e 2]Case 4[β−f(e 3)]+[β+α e 1][β+α(e 1+e 2)]+[β+α e 2][β+α e 2]+[β+α(e 1+e 2)]Case 5[β−f(e 2+e 3)]+[β+α e 1][β−f(e 2)]+[β+α e 1][β−f(e 2)]+[β+α e 1]Case 6[β−f(e 1)]+[β+α(e 2+e 3)][β+α e 3]+[β+α e 2][β+α e 2]+[β+α e 3]Case 7[β+α e 1]+[β+α e 3][β+α(e 1+e 2+e 3)]+[β+α e 2][β+α(e 1+e 2)]+[β+α(e 2+e 3)]