proof of the completion of a certain kind of partial latin square About THE PROOF of the partial latin square has a completion.
Let $r, n \in \mathbb N$, with $r \leq n$. Let $P$ be a PLS of order $n$, in which the ﬁrst $r$ values are used in $n$ entries and the remaining $n − r$ values are not used at all. Then $P$ has a completion.
Can this be proved as a Corollary of the latin rectangle proof by Hall’s Theorem?
 A: Yes it can.  Here's a proof by example:  The partial Latin square
$$L=\begin{array}{|ccccc|}
\hline
\cdot & 1 & \cdot & 2 & \cdot \\
1 & \cdot & \cdot & \cdot & 2 \\
2 & \cdot & \cdot & \cdot & 1 \\
\cdot & \cdot & 2 & 1 & \cdot \\
\cdot & 2 & 1 & \cdot & \cdot \\
\hline
\end{array}$$
is equivalent to the set of entries (called the orthogonal array)
$$\{(1,2,1),(1,4,2),
(2,1,1),(2,5,2),
(3,1,2),(3,5,1),
(4,3,2),(4,4,1),
(5,2,2),(5,3,1)\}.$$
Here, an entry e.g. $(5,3,1)$ says in row 5, column 3, we have symbol 1.
If we swap the first and third coordinates of the elements in the above set, we obtain
$$\{(1,2,1),(2,4,1),
(1,1,2),(2,5,2),
(2,1,3),(1,5,3),
(2,3,4),(1,4,4),
(2,2,5),(1,3,5)\}$$
which is equivalent to the partial Latin square
$$L^{(13)}=\begin{array}{|ccccc|}
\hline
2 & 1 & 5 & 4 & 3 \\
3 & 5 & 4 & 1 & 2 \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\hline
\end{array}.$$  This partial Latin square is called a parastrophe (or conjugate) of the original partial Latin square.
We know $L^{(13)}$ admits a completion by Hall's Theorem, which we denote $(L^{(13)})^c$, and thus $((L^{(13)})^c)^{(13)}$ is a completion of $L$.
To prove this formally, we'd need to show for an arbitrary such partial Latin square:


*

*$L^{(13)}$ is indeed a partial Latin square,

*the first two rows of $L$ are filled, and no other cells are filled,

*$((L^{(13)})^c)^{(13)}$ is a Latin square, and

*$L$ and $((L^{(13)})^c)^{(13)}$ agree in the non-empty cells of $L$.


Each of these checks should take about one sentence.
(Alternatively, we could prove this using Hall's Theorem directly in much the same way as for Latin rectangles.)
