{$\Delta$} Respects Loops:For all {$\langle a,b,c\rangle \in\mathcal {B}^+$}, {$N(\langle a,b,c\rangle )=N(\langle a,b,(a+b+c)\rangle )$}. Proof: The proof that a box {$T$}is loopless iff {$\Delta T$} is loopless shows a correspondence between loops on {$a\times b\times c$} and {$a\times b\times a+b+c$}. The corresponding loops are identical on the base of the box. Since any loop on a box must cross the base, the correspondence includes all the loops on both {$a\times b\times c$} and {$a\times b\times a+b+c$}. |