Every finite matrix over a field is row equivalent to exactly one reduced row echelon matrix

Existence is [L1]. For uniqueness, [L2] applied to $Ax=0$ shows that $R$ and $S$ have the same homogeneous solution set.

The common homogeneous solution set determines the pivot columns. Indeed, column $j$ is a pivot column exactly when every solution $x$ satisfying $x_k=0$ for all $k>j$ also satisfies $x_j=0$. For a pivot column this follows from its pivot equation; for a nonpivot column, set that free variable to $1$ and every other free variable to $0$. Induction over the finite ordered columns therefore gives the same pivot positions for $R$ and $S$.

For each nonpivot column $j$, there is a unique homogeneous solution whose nonpivot coordinates are $1$ at $j$ and $0$ elsewhere. Its coordinate at pivot $p_i$ is the negative of the entry in row $i$, column $j$; pivot columns are unit columns by [L3]. Thus the common solution set determines every entry. With no columns, no rows, or no pivots the same description is empty or all zero, so in every case $R=S$.