Theorem 1 (decomposition theorem)

— Let $(x_i)_{1 \leq i \leq p}$ and $(y_j)_{1 \leq j \leq q}$ be two finite sequences of positive elements of a lattice ordered group $G$ such that

then there exists a double sequence $(z_{ij})_{1 \leq i \leq p, 1 \leq j \leq q}$ of positive elements of $G$ such that $x_i = \sum_{j=1}^q z_{ij}$ for all $i$, and $y_j = \sum_{i=1}^p z_{ij}$ for all $j$.

1) Let us prove the theorem first for the case $p = q = 2$. Let $x, x', y, y'$ be positive elements of $G$ such that $x + x' = y + y'$, and put $a = \sup(0, x - y')$. Since

is less than $x$ and less than $y$, it follows that $b = x - a$ and $c = y - a$ are positive, as is $d = a - (x - y')$. Also we have

2) Let us now show that if the theorem is true for $p < m$ and $q = n$ ($m > 2,\ n \geq 2$) then it is true for $p = m$ and $q = n$. By hypothesis we have $x, + \sum_{i=1}^{m-1} x_i = \sum_{j=1}^n y_j$. The theorem being true for $p = 2$ and $q = n$, there exist two finite sequences $(z_j'), (z_j'')$ of $n$ positive terms such that $\sum_{i=1}^{m-1} x_i = \sum_{j=1}^n z_j'$, $x_m = \sum_{j=1}^n z_j''$, and $y_j = z_j' + z_j''$ for $1 \leq j \leq n$. On the other hand, since the theorem is true for $p = m - 1$ and $q = n$, there exists a double sequence $(u_{ij})_{1 \leq i \leq m-1, 1 \leq j \leq n}$ such that $x_i = \sum_{j=1}^n u_{ij}$ for $1 \leq i \leq m - 1$, and $z_j' = \sum_{i=1}^{m-1} u_{ij}$ for $1 \leq j \leq n$. Putting

we certainly obtain a double sequence satisfying the conditions of the theorem.

3) By interchanging the $x_i$ and the $y_j$ we see in the same way that, if the theorem is true for $p = m$ and $q < n$ ($m \geq 2,\ n > 2$), then it is true for $p = m$ and $q = n$. The theorem is thus proved by double induction starting from the case $p = q = 2$, for it is trivially true whenever $p \leq 1$ or $q \leq 1$.