proof of diamond lemma

Question

I am trying to prove the diamond lemma: Suppose we have two elementary cancellations of a word $w$

then there exists some $w'$ such that there are (possibly trivial) cancellations

The diamond lemma,

There are two cases to consider: whether or not the cancelled subwords in $w$ overlap.

How can I write the proof in the case that the cancelled subwords overlap?

Thank you for your help.

Answer 1

Hint: See Newman's lemma: That is a relation is confluent (satisfies the diamond property) if and only if it is locally confluent.

Answer 2

If the cancellations $aa^{-1}$ and $bb^{-1}$ overlap then either:

1. The occurrences of $aa^{-1}, bb^{-1}$ are the same word, so $w=paa^{-1}q=pbb^{-1}q$ and $w_1=pq=w_2$.
2. If the occurrences of $aa^{-1}, bb^{-1}$ overlap but are non-equal then $w=paa^{-1}aq$, where $a^{-1}a=bb^{-1}$ (i.e. $a=b^{-1}$ and $a^{-1}=b$), and we have $$ paq=w_1\xleftarrow{aa^{-1}}w\xrightarrow{a^{-1}a}w_2=paq. $$

In both cases, we have that $w_1=w_2$, and the result follows.