The link-state form is identically zero only at zero loop value with no defects

Suppose $t>0$ and put $a=(n-t)\mathrm{/}2$. If $a=0$, take $x=y$ to be the all-defect state; gluing gives $t$ through paths and no loops, so ${\varphi }_{n,t}(x,y)=1_R\mathrm{\ne }0$. If $a>0$, let $x$ have adjacent arcs $(1,2),(3,4),\dots ,(2a-1,2a)$ and defects at $2a+1,\dots ,n$. Let $y$ have arcs $(2,3),(4,5),\dots ,(2a,2a+1)$, defects at $1$ and $2a+2,\dots ,n$. Both are link states. Their glued graph has one path alternating across sites $1,\dots ,2a+1$ and $t-1$ straight paths at the shared defects; it has no closed component. Hence ${\varphi }_{n,t}(x,y)=1_R\mathrm{\ne }0$.

Suppose $t=0$ and $n>0$, so $n=2a$ with $a\ge 1$. Every pair of link states glues to a finite graph in which every boundary site has degree two; each connected component is a cycle, and at least one cycle occurs. Thus every form value is divisible by $\delta$, so the form is identically zero when $\delta =0$. Conversely, let $x$ have adjacent arcs $(1,2),(3,4),\dots ,(n-1,n)$, and let $y$ have the outer arc $(1,n)$ together with the adjacent interior arcs $(2,3),(4,5),\dots ,(n-2,n-1)$. Their overlay is one cycle, so ${\varphi }_{n,0}(x,y)=\delta$. If $\delta \mathrm{\ne }0$, this value is nonzero and the form is not identically zero.

If $n=t=0$, there is one empty link state. Its reflection glues to the empty diagram with no loops, so ${\varphi }_{0,0}(x,x)=1_R\mathrm{\ne }0$. Together with the cases already treated, this exhausts all allowed $t$ and proves the equivalence.