Artin's product-cancellation dichotomy

A first middle cancellation must come from original neighbours. Reduce the whole product by deleting adjacent inverse pairs, for example always the leftmost available pair. Before the first middle letter is cancelled, every factor retains its middle letter and therefore has a nonempty residue. Its surviving letters form an interval of the original reduced factor: a deletion inside such an interval is impossible, so deletions take place only at residue boundaries. No factor has disappeared, and these boundaries are between original neighbours. To cancel the left middle letter at the boundary of $T_i,T_{i+1}$, all of $Q_i$ on its right must first cancel against the initial letters of $Q_{i+1}^{-1}$; those letters cannot have been removed at the other boundary without first cancelling the right middle letter. The corresponding assertion holds for cancellation of the right middle letter. Thus any first middle cancellation in the whole product also occurs in a junction cancellation of an original adjacent pair.

If no adjacent pair cancels a middle letter. Step 1.1 shows that no middle letter is cancelled in the whole reduction. All $n$ middle letters therefore survive in its final word $x_1\cdots x_n$ of length $n$, leaving no conjugator letters and forcing their order to be $x_1,\dots ,x_n$. The initial $Q_1^{-1}$ cannot be deleted: it is reduced, has no factor on its left, and cannot cancel across its surviving middle letter. Hence $Q_1$ is empty. Inductively, if $Q_1,\dots ,Q_{j-1}$ are empty, the initial letters of $Q_j^{-1}$ cannot cancel against the surviving earlier middle letters or across its own middle letter. Hence $Q_j$ is empty as well. Thus all conjugators vanish and $\pi (j)=j$, so $A=\mathrm{id}$. This includes $n=0,1$.

The two explicit prefix forms. Otherwise choose the least qualifying $i$. At that junction the words are $U^{-1}aU$ and $V^{-1}bV$. If $a$ is the first middle letter cancelled, cancellation removes the entire suffix $U$ of the left factor against the head $U^{-1}$ of $V^{-1}$, and the next letter of $V^{-1}$ must be $a^{-1}$. Equivalently $V=RaU$ as a reduced word. If $b$ is the first cancelled middle letter, the whole head $V^{-1}$ has cancelled against the suffix $V$ of $U$, and the preceding letter of $U$ must be $b^{-1}$; hence $U=R{b}^{-1}V$. The two positive middle letters cannot cancel each other, so the first deletion involves exactly one of them. Exhaustiveness and exclusivity follow by whether a qualifying junction exists.