The paper's reduction of the equality problem to commutativity. $g$ is supported only on words beginning $fresh0 · fresh1$, where $g (fresh0 · fresh1 · w) = f w$. If $g$ is commutative, then $f w = g (fresh0 · fresh1 · w) = g (fresh1 · fresh0 · w) = 0$ (the two words are commutatively equivalent), so $f = 0$. Conversely $f = 0$ forces $g = 0$, which is commutative.