Composition transfers a constant fraction of unsatisfaction

If $E(G)=\mathrm{\varnothing }$, then $\mathrm{UNSAT}(G)=0$ by [F6]. The composition has an empty constraint list by [F8], so $\mathrm{UNSAT}(H)=0$ by [F7]. The claimed global inequality follows.

Suppose $E(G)\mathrm{\ne }\mathrm{\varnothing }$. For each active vertex $v$, let $B_v$ be its physical block under $\tau$ and decode it by the nearest-codeword rule of [F3]. Each vertex has one shared block by [F4], so this gives one label at that vertex for every incident edge, including both formal positions of a loop. Assign the first symbol of $\mathrm{\Sigma }$ to any isolated vertex; by [F6] this extension does not change graph unsatisfaction. Denote the resulting global graph labeling by ${\sigma }_{\tau }$.

For each edge $e$, let ${\tau }_e$ be the pullback of $\tau$ to its local gadget under the composition map. The equal-row construction in [F5] gives ${\mathrm{UNSAT}}_{\tau }(H)=\frac{1}{\mathrm{\mid }E(G)\mathrm{\mid }}{\sum }_{e\in E(G)}{\mathrm{UNSAT}}_{{\tau }_e}(T_e)$.

Let $F_{\tau }$ be the edge records violated by ${\sigma }_{\tau }$. For each $e=(v,w)\in F_{\tau }$, the named input to its local tester is the formal pair $(B_v,B_w)$, with $(B_v,B_v)$ for a loop. By [F3] this input is at relative distance at least $\delta \mathrm{/}4$ from the edge circuit's accepting set. By [F2] and [F4], the local gadget is the assignment tester for that circuit on both named pieces; applying [F1] to the pullback labeling ${\tau }_e$ therefore gives ${\mathrm{UNSAT}}_{{\tau }_e}(T_e)\ge {\rho }_0\delta \mathrm{/}4$. For an edge outside $F_{\tau }$, its local unsatisfaction is at least zero by [F7].

Combine the identity of step 1.3 with the local bounds of step 2.1. Since $\mathrm{\mid }{F}_{\tau }\mathrm{\mid }\mathrm{/}\mathrm{\mid }E(G)\mathrm{\mid }={\mathrm{UNSAT}}_{{\sigma }_{\tau }}(G)$ by [F6], and $\mathrm{UNSAT}(G)$ is the minimum over graph labelings, it follows that ${\mathrm{UNSAT}}_{\tau }(H)\ge \frac{{\rho }_0\delta }{4}\frac{\mathrm{\mid }{F}_{\tau }\mathrm{\mid }}{\mathrm{\mid }E(G)\mathrm{\mid }}=\frac{{\rho }_0\delta }{4}{\mathrm{UNSAT}}_{{\sigma }_{\tau }}(G)\ge \frac{{\rho }_0\delta }{4}\mathrm{UNSAT}(G)$.

Step 3.1 holds for every Boolean labeling $\tau$ of the finite output system. Taking the minimum over those labelings gives the asserted inequality for $\mathrm{UNSAT}(H)$. The edgeless case was handled in step 1.1, and ${\rho }_0\delta \mathrm{/}4=(1\mathrm{/}1000)(1\mathrm{/}2)\mathrm{/}4=1\mathrm{/}8000$.