Repeating constraints amplifies the gap

Let $G$ have $M$ constraints and let $G^{(r)}$ be the system obtained by listing every constraint of $G$ exactly $r$ times, $r\ge 1$. If $M=0$, both lists are empty and the value convention of [F1] gives unsatisfaction zero for every labeling in both systems. If $M>0$, fix a labeling $\sigma$ and let $v$ be the number of constraints of $G$ violated by $\sigma$: the list $G^{(r)}$ has $rM$ constraints and exactly $rv$ of them are violated, so ${\mathrm{UNSAT}}_{\sigma }(G^{(r)})=rv\mathrm{/}(rM)=v\mathrm{/}M={\mathrm{UNSAT}}_{\sigma }(G)$; taking the minimum over labelings gives $\mathrm{UNSAT}(G^{(r)})=\mathrm{UNSAT}(G)$, never a strict increase.

The failure is nonvacuous. Take the alphabet $\mathrm{\Sigma }:=\{0,1\}$, the one-vertex constraint graph $V:=\{x_1\}$ with the two loops of Constraint graph and labeling value: the loop carrying the relation $R_1:=\{(0,0),(1,1)\}$, which every labeling of $x_1$ satisfies, and the loop carrying the empty relation $R_2:=\mathrm{\varnothing }$, which every labeling violates. Then $G$ has $M=2$ constraints, every labeling violates exactly one of them, and $\mathrm{UNSAT}(G)=1\mathrm{/}2$; after repeating each constraint $r$ times every labeling violates exactly $r$ of the $2r$ listed constraints, so $\mathrm{UNSAT}(G^{(r)})=1\mathrm{/}2$ for every $r\ge 1$.

So the false statement fails at the witness of step 1.2 for every $r\ge 1$, and by step 1.1 no instance whatsoever has its unsatisfaction fraction changed by an equal repetition of its constraints. What an amplification step needs instead is a change of the variables and constraints, not a reweighting of the existing list: the powering of Constraint graph powering with local-view labels and the comparison families of the assignment-tester construction are of this genuinely different kind.