I'm reading a PDF about completing partial latin squares. The reference is https://ajc.maths.uq.edu.au/pdf/22/ocr-ajc-v22-p247.pdf
There is a fact that doesn't have a proof here, and I can't find it:
The back diagonal of a partial latin square $P$ of order $n$ is formed by the cells ${(1, n), (2, n-1), ... , (n, 1)}.$ An entry $(x, y, s)$ of $P$, is said to lie on the back diagonal of $P$ if $(x, y, s) \in P$ and $x + y = n + 1$, $(x, y, s)$ lies above the back diagonal if $(x, y, s) \in P$ and $x + y < n + 1$, and $(x, y, s)$ lies below the back diagonal if $(x, y, s) \in P$ and $x + y > n + 1$. Let $L$ be a latin square of order $n$, then $L$ can be used to construct a new partial latin square $P( L)$ as follows:
- If $(x, y, s)$ lies on or above the back diagonal in $L$, then $(x, y, s) \in P(L)$.
- For all $i = 1, ... , n + 1, (i, n + 2 - i, n + 1) \in P (L ) $.
Let $L$ be any latin square, of order $n$. The partial latin square $P(L)$, of order $n + 1$ (as defined above), can always be completed.
Can anyone help me with the proof?