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What {$\Delta$} Does:

For {$T\in \mathcal {B}^+$}, {$k\in\mathbb N$},

1. {$\Delta^k(T)$} has no more than one negative member
2. {$\Delta^k(T)$} has at least one positive member
3. {$S(\Delta^k(T))>0$}, where {$S(T)$} is the sum of the members of {$T$}
4. If {$\Delta^k(T)$} has a negative member, it is least in absolute value.

Proof:

These all follow inductively. For 1., {$\Delta(T)$} can only have one more negative or non-positive member than {$T$}. If {$\Delta(t)=\{a,b,c\}$}, {$a<0\le b\le c$}, then {$c-b\ge 0$} so {$c-b-a>0$}.

For 2., if {$\Delta(T)=\{a,b,c\}$}, {$a,b\le 0<c$}, then {$c-a-b>0$}.

For 3., if {$\Delta(T)=\{a,b,c\}$}, {$a\le b\le c$} then {$c>0$} by (ii), so {$a+b+(c-a-b)=c>0$}.

For 4., suppose first that {$j$} is greatest such that {$\Delta^j(T)=\{a,b,c\}$}, has no negative members, {$a,b\le c$}, so {$c-a-b<0$} and the absolute value of the negative member of {$\Delta^{j+1}(T)$} is {$a+b-c$}. From {$b\le c$} and {$a\le c$} it follows that {$a+b-c\le a$} and {$a+b-c\le b$}, respectively.

Now suppose that {$\Delta^j(T)=\{a,b,c\}$} with {$c<0\le a\le b$} and 4. holds. Then 4. continues to hold for {$\Delta^{j+1}(T)$} since {$-c\le b-(a+c)$} follows from {$a\le b$}.