then $w'' = w\wedge w'\in \mathbb{L}_{2n}^N$. We need to show that $w'' = (w''_1,\ldots,w''_{2n})$ satisfies conditions (i)-(iv) of Definition \[def:calL\]. Since $w_{2k-1} \leq w_{2k}$ and $w'_{2k-1} \leq w'_{2k}$ $$w''_{2k-1}:= \min\{w_{2k-1},w'_{2k-1}\} \leq \max\{w_{2k},w'_{2k}\} = w''_{2k}$$ and thus (i) is satisfied. To see that (ii) is satisfied we note that $w_{2k-1} < w_{2k+1}$ and $w'_{2k-1} < w'_{2k+1}$ implies that $$w''_{2k-1}:= \min\{w_{2k-1},w'_{2k-1}\} < \min\{w_{2k+1},w'_{2k+1}\} = w''_{2k+1}.$$ Similar arguments apply to (iii) and (iv).

Checking that $\wedge$ is associative, symmetric, and idempotent is straightforward.

We denote the set of *ordered multi-indices* by $$\label{ordmultind}
\mathcal{L}_{2n}^N := \left\{ ( x_1,\dots,x_{2n} ) \in \{ 1,\dots,N\}^{2n}\colon x_{2k-1}\leq x_{2k}\text{ for $1\leq k \leq n$ and }x_{2k}<x_{2k+1}\text{ for $1\leq k \leq n-1$} \right\}$$ Observe that $\mathcal{L}_{2n}^N\subset \mathbb{L}_{2n}^N$. However, $\mathcal{L}_{2n}^N$ is not a sub meet-semilattice of $\mathbb{L}_{2n}^N$. To see this assume that $x,x'\in \mathcal{L}_{2n}^N$ and that $x'_{2k} < x'_{2k+1} < x_{2k} < x_{2k+1}$. Let $x'' = x'\wedge x$. Then $x''_{2k} = x_{2k} \not< x'_{2k+1} = x''_{2k+1}$, which implies that $x''\not\in\mathcal{L}_{2n}^N$.

\[ex:calL34\] The simplest nontrivial example of a set of ordered multi-indices is $$\begin{aligned}
\mathcal{L}_{4}^3 & = {\left\{ {(x_1,x_2,x_3,x_4) : x_1\leq x_2 < x_3\leq x_4}\right\}} \\
& = {\left\{ {(1122),(1123),(1133),(1233),(2233)}\right\}}\end{aligned}$$

Recall the strictly ordered multi-indices $L_{2n}^N$ defined in Section \[sec:C(h)\] and observe 