The finite-table coloring algorithm terminates and is correct

Let $s$ be the number of rows in the candidate. The shape check reads the row count and all $s$ row lengths, so it terminates after $s+1$ shape checks and returns that witness if the shape fails. If the shape is valid, then $s=m$; the entry-range check examines at most $m^2$ values, and the column-permutation checks examine at most $m^2$ entries. Thus shape and format validation use at most $m+1+2m^2$ checks on a shape-valid table, and a malformed shape stops before any entry or axiom checks.

The triple loop visits exactly $m^3$ ordered triples and checks the right self-distributive identity of [F2]. In quandle mode the diagonal loop visits $m$ entries and checks exactly $T[x,x]=x$. Thus the total check bound on a shape-valid table is at most $(m+1)+2m^2+m^3+m=m^3+2m^2+2m+1$. The algorithm accepts exactly the tables satisfying the selected rack axioms, with idempotency required precisely in quandle mode.

If a column passes its permutation check, it defines a bijective right translation and its inverse column is completely determined. The assignment loop enumerates every one of the $m^a$ functions from the ordered arc set to the table elements once. At each crossing it tests exactly the relation from [F2], using that inverse column if required; hence an assignment is retained if and only if it is a coloring.

Every loop has a finite range, so the procedure terminates; it returns the exact failure witness at the first failed check or the complete coloring set and count when all checks pass. For the empty diagram the unique empty assignment is tested once, and for a diagram with zero crossings every one of the $m^a$ assignments passes.