 Solution

Read the arrows upwards through the central part of the diagram. The lower anyon fusion tree first splits $a$ into $b,c$, then $b$ into $d,j$ and $c$ into $h,e$. The upper anyon fusion tree combines $d,j$ into $f$, $h,e$ into $g$, and finally $f,g$ into $a$. The two interior crossings are not vertices. Thus admissibility requires the six anyon fusion multiplicitiesIf any factor vanishes, the diagram has zero value.

Make an anyon vertex gauge transformation by multiplying each splitting vertex $z\to x,y$ by an independent wave phase $u_z^{xy}$ and its adjoint fusion vertex by $(u_z^{xy}{)}^{-1}$. The diagram changes byTherefore the label-matching condition for cancellation of all anyon vertex gauge transformation factors iswith $a,d,e,h,j$ otherwise unrestricted apart from the anyon fusion rules. This statement also allows coincident labels: the three ordered vertex triples in the numerator must equal those in the denominator as multisets; matching the triples in any of the six possible orders still forces $b=f,c=g$. Conversely, these equalities visibly give $G=1$.

Closure of the drawing alone does not guarantee invariance under an anyon vertex gauge transformation: unmatched fusion and splitting vertices retain their independently chosen basis phases. A zero value is, of course, invariant regardless of phases; the boxed condition describes cancellation of the vertex factors themselves.