One Dinur transformation preserves perfect satisfiability

Assume first that $E(G)=\mathrm{\varnothing }$. Then $\mathrm{val}(G)=1$ by [F4]. The completeness clause of [F2] applied to this input gives $\mathrm{val}(R_t(G))=1$, and $T_t(G)$ is edgeless by [F1] and [F2]. Applying the forward value-one direction of [F3] at $\mathrm{\Sigma }={\mathrm{\Sigma }}_t$ to the graph $R_t(G)$ therefore gives $\mathrm{val}(T_t(G))=\mathrm{val}(A_{{\mathrm{\Sigma }}_t}(R_t(G)))=1$.

Assume now that $E(G)\mathrm{\ne }\mathrm{\varnothing }$. The identity $T_t(G)=A_{{\mathrm{\Sigma }}_t}(R_t(G))$ of [F1] rewrites the goal as $\mathrm{val}(A_{{\mathrm{\Sigma }}_t}(R_t(G)))=1$. The intermediate graph $R_t(G)$ is a finite binary constraint graph over the alphabet ${\mathrm{\Sigma }}_t$ supplied by [F2] and [F1].

The completeness clause of [F2] applies to the input $G$ because $\mathrm{val}(G)=1$, so $\mathrm{val}(R_t(G))=1$.

Apply the forward direction of [F3] with the fixed alphabet ${\mathrm{\Sigma }}_t$, whose size is at least two, to the graph $R_t(G)$. Since $\mathrm{val}(R_t(G))=1$, it gives $\mathrm{val}(A_{{\mathrm{\Sigma }}_t}(R_t(G)))=1$, that is, $\mathrm{val}(T_t(G))=1$ by the identity of [F1].

Step 1.1 proves the edgeless case and step 2.1 proves the case of a nonempty edge set; these two cases exhaust all finite inputs, so for the arbitrarily fixed $t\ge t_0$, $\mathrm{val}(G)=1$ implies $\mathrm{val}(T_t(G))=1$. No random sampling or selection from a varying family occurs: the two maps are deterministic and the cases are decided by whether the finite edge set is empty.