Last week I was recommended Bergman's Diamond Lemma in these comments. I read through the paper, and was working on an exercise in it on page 193:

Examine for termination each of the following singleton reduction-systems on $\mathbf{k}\langle x, y \rangle$: $\{(x^2y, yx)\}$, $\{(yx, x^2y)\}$, $\{(x^2y^2, yx)\}$, $\{(yx, x^2y^2)\}$.

I want to know if the reduction systems $\{(x^2y^2,yx)\}$ and $\{(yx,x^2y^2)\}$ lead to a normal form.

  • The rule $(x^2y^2,yx)$ is length reducing and has no overlap ambiguities, and so $(x^2y^2,yx)$ leads to a normal form.

  • On the other hand, the rule $(yx,x^2y^2)$ has no overlaps. In this case, consider the term $y^2x$. Under this reduction, we have the series of reductions \begin{gather*} y^2x \\ yx^2y^2 \\ x^2y^2xy^2\\ x^2yx^2y^2y^2\\ x^2x^2y^2xy^4\\ x^4yx^2y^2y^4\\ x^4yx^2y^6\\ x^4x^2y^2xy^6\\\vdots \end{gather*} and evidently this term does not reduce to a normal form as there is always some $x$ to the right of some $y$ after performing a reduction.

Did I understand these correctly? I wasn't sure about the second case, since I felt lucky that I chose such a monomial to reduce.


1 Answer 1


Yeah, you're right. And it's not just a lucky guess of word $y^2x$; it's the perfect example of a word that's not reduction-finite, so you know that the rule $\{(yx,x^2y^2)\}$ won't terminate for every word in $\mathbf{k}\langle x,y \rangle$, and so that rule doesn't give you a normal form. All you need is that one counterexample. :)


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