# Sudoku - good proof of one entry must be 1

The question is from here

Let the set of numbers $N = \{1, 2,\ldots , 9\}$. Let the set of cells $C = N × N$.

A Sudoku solution $f$ is a function $f : C \to N$ satisfying the following properties:

• For all $i, j,$ and $j′,$ if $j ≠ j′$ then $f(i, j)≠f(i, j′)$
• For all $i, i′,$ and $j,$ if $i ≠ i′$ then $f(i, j)≠f(i′,j)$
• For all $c1$ and $c2 \in C$, if $c1 ≠ c2$ and $square(c1)=square(c2)$, then $f(c1)≠f(c2)$.

$square(i, j)$ is the number of the large box that contains the $(i, j)$ cell, where we number the large $9x9$ grids as follows:

1 4 7
2 5 8
3 6 9

The claim is: if f is a sudoku solution compatible with the board below (that is, if $f(2, 2)=1, f(1, 5)=8, f(2, 5)=5$ and so on), then $f(9, 3)=1$. And the given proof is:
There must be a $1$ somewhere in box $3$ (in other words, there exists some $(i, j)$ such that $square(i, j)=3$ and $f(i, j)=1$). By rule $2$, $j$ cannot be $1$, because $f(4, 1)=1$. Similarly, since $f(2, 2)=1$, $j$ cannot be $2$. Thus $j$ must be $3$; and $f(7, 3)\ne 1$, and $f(8, 3)\ne 1$. The only cell left is $(9, 3)$, so $f(9, 3)$ must be $1$.

It is suggested that it's not a good proof. But why? And what should be a good proof like to this problem?

There must be a $1$ somewhere in box $3$
which has not yet been proven. It's "obvious", but not quite obvious enough to simply state it. And that result is going to be useful, so worth proving out in a general way that for any $k, m \in N$, there must be some $c \in C$ such that $square(c) = k$ and $f(c)=m$
Then the discussion of the other entries with a $1$ in the first two columns needs to mention that those entries are not in square $3$. Again "obvious" but in order to use the rules as stated, that should be pointed out, perhaps by noting that $square(i,j)=3 \implies i \in \{1,2,3\}$ and $j\in \{7,8,9\}$