Proof that the incidence matrix of a laminar family is TU. A friend and I wrote a proof for this using the consecutive ones property that I haven't seen anywhere, so I thought I would share it here.
$\textit{Def:}$ A matrix has the Total-Unimodularity (TU) property if the determinant of all of its square submatrices has determinant 1, -1, or 0.
$\textit{Def:}$ A family of subsets $S$, is called laminar if for all $s,t \subset S$, if $s \cap t$ is not empty, then either $s \subset t$, or $s \supset t$.
Let $S$ be a laminar family of a nonempty finite set $V$, and let $A_5$ denote its $M \times N$ incidence matrix. 
Prove that $A_S$ is TU.
 A: $\textbf{Claim:}$ $A_S$ can be permuted into a matrix having the consecutive ones property.
$\textbf{Proof:}$ Permute the rows of $A_S$ so that they have descending row-wise sums.
$\textit{Base Step:}$ Permute the columns so that the first row has the consecutive ones property.
$\textit{Inductive Step:}$ Assume the first $N-1$ rows have the consecutive ones property. Then consider the $N^{th}$ row, $r_l$, and an arbitrary row $r_u$ from the first $N-1$ rows. 
For the inductive step it will suffice to show that we can permute the columns of our permuted $A_S$ matrix to ensure the consecutive 1's property in $r_l$, while maintaining this property in $r_u$.
By the Laminar Property and the ordering of the rows of $A_S$, the set $l^{th}$ set in $S$, $s_l$ has either $s_l \subset s_u$, or $s_l \cap s_u = \emptyset.$ Thus, over all the indices $i$ where $r_l$ is 1, the value of $r_u$ is either all 1, or all 0. By the consecutive ones property, all of the indices within this selection are the same value of 1, or all 0. 
Then we can freely permute the columns where $r_l$ is 1, and all of the ones within this selection, without altering the values in any of the rows above $r_l$. Thus, we can permute this subset of columns of $A_S$ to give $r_l$ the consecutive ones property, without compromising the consecutive ones property of any of the first $N-1$ rows.
$\textbf{Note:}$ As shown here, $M$ has the consecutive 1's property $\Rightarrow$ $M$ is TU. In addition, all permutations of $M$ are also TU.
