A single equality edge through robust composition

By [F1] the dimension-one table is evaluated at masks $0$ and $1$, so ${\mathrm{WH}}_1(1)=(1\cdot 0,1\cdot 1)=(0,1)$ with length $\mathrm{\ell }=2^1=2$, in agreement with the code selection of [F2].

The labeling $\sigma (v)=\sigma (w)=1$ is a graph labeling, and its ordered endpoint pair $(1,1)$ lies in $R_e=\{(0,0),(1,1)\}$; since the equality edge is the only edge, ${\mathrm{val}}_{\sigma }(G)=1$ and $\mathrm{val}(G)=1$, so $\mathrm{UNSAT}(G)=0$ by [F3].

The block assigned to both endpoints is $B_v=B_w=C(1)=(0,1)$ by [F2]. Both blocks are valid codewords and decode uniquely to the labels $1,1$, whose ordered pair lies in $R_e$; hence the robust edge circuit accepts the displayed input by [F2].

Since $\mathrm{val}(G)=1$, [F4] gives $\mathrm{val}(H)=1$: some assignment to the variables of $H$ satisfies every constraint of $H$. By [F5], the single edge contributes $M=q_e$ constraints, namely the constraints of the local two-piece tester copied once each; therefore one assignment satisfies every tester gadget constraint simultaneously.

Apply the conversion of [F6] to $H$. The satisfying assignment of step 2.2 extends to a labeling of the output graph that satisfies every output edge, so the output has value one and unsatisfaction zero: the number of violated edges is $0$. The conversion creates at most $6M$ edge records and uses the fixed 66-symbol alphabet $\{B(0),B(1)\}\bigsqcup \{0,1{\}}^6$, so this is a concrete instance of the composition and conversion maps with no random or infinite selection anywhere.