All Terminate:Every {$T\in \mathcal{B}^+$} is terminating. Proof: Consider {$\{S(\Delta^k(T))\}_n$}. Clearly this is a non-increasing sequence. It follows from Proposition 2, then, that at some point, {$S(\Delta^{k-1}(T))=S(\Delta^k(T))$}. Then if {$a\le b\le c$} are the members of {$\Delta^{k-1}(T)$}, then {$S(\Delta^k(T))=a+b+(c-a-b)$} and so {$a+b=0$}. Then {$\Delta^{k-1}(T)=\Delta^k(T)$} and {$T$} is terminating. |