Successive application of binary operation to set of n-tuples produces a semigroup. Because I am not a mathematician the initial versions of this question will not as precise as they need to be. I therefore expect to refine and clarify question based on comments and feedback. Thanks for your patience.
Given a set $A$ of $n$-tuples, for example 
$$
[ (3, 5, 5), (5, 3, 4), (1, 6, 3), (2, 1, 6), (6, 2, 1), (4, 4, 2) ]
$$
where $n = 3$, and a binary operation $\land$ such that for any given pair of tuples $(x_1 , y_1 , z_1)$ and $(x_2 , y_2 , z_2)$
$$
(x_1 , y_1 , z_1) \land (x_2 , y_2 , z_2) = ( min(x_1,x_2) , min(y_1,y_2) , min(z_1,z_2) )
$$
Applying $\land$ to all pair of tuples in the set $A$ yields the set $B$ (ignoring duplicates)
$$
[(1, 2, 1), (1, 5, 3), (2, 1, 2), (3, 2, 1), (2, 1, 4), (2, 1, 1), (4, 3, 2), (2, 1, 5), (1, 1, 3), (3, 3, 4), (1, 4, 2), (4, 2, 1), (3, 4, 2), (1, 3, 3), (5, 2, 1)]
$$
Applying $\land$ again to all pair of tuples in the set $B$ yields the set $C$ (again ignoring duplicates)
$$
[ (1, 2, 1), (2, 1, 1), (1, 3, 2), (3, 3, 2), (2, 1, 4), (1, 4, 2), (3, 2, 1), (4, 2, 1), (1, 1, 2), (1, 1, 3), (2, 1, 2), (1, 1, 1), (1, 3, 3) ]
$$
The set $C$ has an interesting property that I've observed. Namely subsequent applications of $\land$ to any pair of tuples in $C$ do not generate any new tuples outside the set.
More generally, I've observed that for any set of $n$-tuples, $n - 1$ successive applications of $\land$ generates a set of $n$-tuples that is closed under $\land$. To my limited understanding $C$ is a semigroup, where $A$ and $B$ are semigroupoids.
My real question though is why do subsequent applications of the binary operation $\land$ to a semigroupoid result in a semigroup?
 A: This is a nice question, which is actually a problem on semilattices.
Let $\mathbb{N}$ be the semigroup of natural numbers under the operation
$x \wedge y = \min(x, y)$. This semigroup is clearly commutative but is also idempotent, that is, $x \wedge x = x$ for all $x \in \mathbb{N}$. Commutative and idempotent semigroups are also called semilattices.
Now, the product of $k$ copies of $(\mathbb{N}, \wedge)$ yields the semilattice $(\mathbb{N}^k, \wedge)$ with operation defined componentwise. That is, if $x = (x_1, \ldots, x_k)$ and $y = (y_1, \ldots, y_k)$ then
$$
  x \wedge y = (x_1, \ldots, x_k)(y_1, \ldots, y_k) = (x_1 \wedge y_1, \ldots, x_k \wedge y_k).
$$
If $S$ is a subset of $\mathbb{N}^k$, let us define $S^n$ by induction by setting $S^1 = S$ and $S^{n+1} = \{s \wedge t \mid s \in S \text{ and } t \in S^n\}$. Now your result can be stated as follows:

For every $k > 0$, the set $S^k$ is equal to the subsemigroup of
  $\mathbb{N}^k$ generated by $S$.

Before proving the general case, it is interesting to first look at the case $k = 2$, which already contains the main argument of the proof. An element of the subsemigroup of $\mathbb{N}^2$ generated by $S$ is of the form 
$$
  t=(x_1, y_1) \wedge (x_2, y_2) \wedge  \dotsm  \wedge (x_n,y_n) = (\min\{x_1, \ldots, x_n\}, \min\{y_1, \ldots, y_n\})
$$
with $(x_1, y_1), (x_2, y_2), \dotsm, (x_n,y_n) \in S$. The problem is to show that $t$ belongs to $S^2$. Let $i$ and $j$ be such that
\begin{align}
 (1) \quad x_i &= \min\{x_1, \ldots, x_n\} \\
 (2) \quad y_j &= \min(\{y_1, \ldots, y_n\} - \{y_i\}).
\end{align} 
I claim that 
$$
t = (x_i, y_i) \wedge (x_j, y_j) = (\min\{x_i, x_j\}, \min\{y_i, y_j\}).
$$
Indeed, by (1), $$\min\{x_i, x_j\} = x_i = \min\{x_1, \ldots, x_n\}$$ and by (2)
$$\min\{y_i, y_j\} = \min(y_i, \min(\{y_1, \ldots, y_n\} - \{y_i\}) = \min(\{y_1, \ldots, y_n\}$$
which proves the claim and shows that $t \in S^2$.
The general case is not much more difficult, but requires double indices. Let $S$ be a subset of $\mathbb{N}^k$. An element of the subsemigroup of $\mathbb{N}^k$ generated by $S$ is of the form 
$$
  t= x_1 \wedge x_2 \wedge  \dotsm  x_n 
$$
where $x_1 = (x_{1,1}, \ldots, x_{1,k}), \dots, x_n = (x_{n,1}, \ldots, x_{n,k}) \in S$. Let us show that $t \in S^k$. Let $i_1, \ldots, i_k$ be such that
\begin{align}
 x_{i_1,1} &= \min\{x_{1,1}, \ldots, x_{n,1}\} \\
 x_{i_2,2} &= \min(\{x_{1,2}, \ldots, x_{n,2}\} - \{x_{i_1,2}\}) \\
\vdots   &=\vdots\\
 x_{i_k,k} &= \min(\{x_{1,k}, \ldots, x_{n,k}\} - \{x_{i_1,k}, \ldots, x_{i_{k-1},k}\})\\
\end{align} 
I let you verify that $t = x_{i_1} \wedge \dotsm \wedge x_{i_k}$ and thus belongs to $S^k$.
