And True Inductively Claim:Let {$P^1$} and {$P^3$} be paths on {$T=\langle r,s,t\rangle$} corresponding to {$L_1$} and {$L_3$}. If the statements
are true on {$T$}, then they are true on all {$\Delta$} pre-images of {$T$}. Proof: We won't do all three here; they are similar. We will address the second pre-image. We assume that the three statements are true of {$T=\langle r,s,t\rangle$}. Let {$T'=\langle r,r+s+t,t\rangle$}, the second pre-image. With the linear transformation, {$$K(u,v,w)=(v+w,u+w,u+v),$$} the three equations of our assumption can be rewritten: {$$\text{GCD}_{_T}\left[\begin{array}{ccc}p^1_a&p^1_b&p^1_c\\ p^1_a+p^3_a&p^1_b+p^3_b&p^1_c+p^3_c\end{array} \right]=K(T).$$} With the transformations, {$$J(u,v,w)=(u+v,v,w+v)$$} {$$H(u,v,w)=(u,u+v+w,w).$$} {$H(T)$} is {$T'$}, the second pre-image and {$J(P^1)$} is the path vector of the path on {$T'$} corresponding to {$P^1$}, so what we need to show is {$$\text{GCD}_{_{T'}} J\left(\left[\begin{array}{ccc}p^1_a&p^1_b&p^1_c\\ p^1_a+p^3_a&p^1_b+p^3_b&p^1_c+p^3_c\end{array} \right]\right)=K(H(T)),$$} where {$J$} is applied to each column. But GCD{$_{_{T'}}=$}GCD{$_{_{T'}}$} and {$$\text{GCD}_{_T} J\left(\left[\begin{array}{ccc}p^1_a&p^1_b&p^1_c\\ p^1_a+p^3_a&p^1_b+p^3_b&p^1_c+p^3_c\end{array} \right]\right)=J \left(\text{GCD}\left[\begin{array}{ccc}p^1_a&p^1_b&p^1_c\\ p^1_a+p^3_a&p^1_b+p^3_b&p^1_c+p^3_c\end{array} \right] \right)=J(K(T)),$$} so we need only show that {$J(K(T))=K(H(T))$}. But this is easily done: {$$\begin{array}{rcl}J(K(T))&=&J(s+t,r+t,r+s)\\ &=&((s+t)+(r+t),r+t,(r+s)+(r+t))\\ &=&((r+s+t)+t, r+t, r+(r+s+t))\\ &=&K(r, r+s+t, t)\\ &=&K(H(T)).\end{array}$$} |