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\begin{proof}[Proof of \cref{thm:3.13}]
  By inductionon $m = \text{number of $\nu_{-1}, \ldots, \nu_{-4}$ that are not
  identically $1$}$.
  $m = 0$ and $m = 1$ straightforward.

  Suppose true for $m = M \ge 1$, and consider $M + 1$.
  Without loss of generality $\nu_{-1}$ is not identically $1$.
  Define auxiliary weighted hypergraphs $\nu', g', \tilde{g}' : X_{-1} \to [0,
  \infty)$ by
  \begin{align*}
    \nu'(x_{-1})
    &= \Ebb_{x_1 \in X_1} \nu_{-2}(x_{-2}) \nu{-3}(x_{-3}) \nu_{-4}(x_{-4}) \\
    g'(x_{-1})
    &= \Ebb_{x_1 \in X_1} g_{-2}(x_{-2}) g_{-3}(x_{-3}) g_{-4}(x_{-4}) \\
    \tilde{g}'(x_{-1})
    &= \Ebb_{x_1 \in X_1} \tilde{g}_{-2}(x_{-2}) \tilde{g}_{-3}(x_{-3})
    \tilde{g}_{-4}(x_{-4})
  \end{align*}
  Note that these do not depend on $\nu_{-1}$, $g_{-1}$, $\tilde{g}_{-1}$.
  Also define $g' \wedge 1 = \max \{g', 1\}$ and $\nu' \wedge 1 = \max \{\nu',
  1\}$.

  Observe that by the LFC, $\Ebb \nu', \Ebb \nu'^2 = 1 + o(1)$ and by
  Cauchy--Schwarz, $(\Ebb |\nu' - 1|)^2 \le \Ebb |\nu' - 1|^2 = o(1)$ (i).

  \textbf{Claim:} $\|g' \wedge 1 - \tilde{g}'\|_{\square} = o(1)$.

  Proof of claim: Since $0 \le g' - g' \wedge 1 = \max \{g' - 1, 0\} \le \max
  \{\nu' - 1, 0\} \le |\nu' - 1|$ (ii).
  For any $A_2 \subseteq X_3 \times X_4$, $A_3 \subseteq X_2 \times X_4$, $A_4
  \subseteq X_2 \times X_3$,
  \begin{align*}
    &\!\!\!\!
    \Ebb((g' \wedge 1) - \tilde{g}'(x_{-1})) \ub{\indicator{A_2}(x_3, x_4)
    \indicator{A_3}(x_2, x_4) \indicator{A_4}(x_2, x_3)}_{= \prod_{j = 2}^4
    \indicator{A_j}} \\
    &= \ub{\Ebb (g' \wedge 1 - g')(x_{-1})}_{\text{in absolute value $= o(1)$
    because of (i) and (ii)}} \prod_{j = 2}^4 \indicator{A_j}
    + \Ebb (g' - \tilde{g}') (x_{-1}) \prod_{j = 2}^4 \indicator{A_j}
  \end{align*}
  Note that $g'$ involves at most $M$ unbounded functions, so the second term is
  $o(1)$ by the inductive hypothesis.
\end{proof}

Consider
\begin{align*}
  \left| \Ebb \prod_{j = 1}^{4} g_{-j}(x_{-j}) - \prod_{j = 1}^{4}
  \tilde{g}_{-j}(x_{-j}) \right|
  &= |\Ebb_{x_2, x_3, x_4} g_{-1}(x_{-1}) g'(x_{-1}) - \tilde{g}_{-1}(x_{-1})
  \tilde{g}'(x_{-1})| \\
  &= \Ebb g_{-1} g' - \tilde{g}_{-1} \tilde{g}' \\
  &= \Ebb g(g' - \tilde{g}')+ \Ebb (g - \tilde{g}) \tilde{g}'
\end{align*}
Since $0 \le \tilde{g} \le 1$, the second term is at most $\|g -
\tilde{g}\|_{\square} = o(1)$.
The first term, by Cauchy--Schwarz, is bounded above by
\[
  (\Ebb g) (\Ebb g(g' - \tilde{g}')^2)
  \le \ub{\Ebb \nu}_{= 1 + o(1)} \cdot \Ebb \nu(g' - \tilde{g}')^2
\]
By the strong linear forms condition, may replace $\nu$ by $1$.
So $|\Ebb g(g' - \tilde{g}')|^2 \le (1 + o(1))(o(1) + \Ebb(g' - \tilde{g}')^2)$.
Expand
\[
  \Ebb(g' - \tilde{g}')^2
  = \Ebb (\ub{g'}_{\le \nu'} - \ub{\tilde{g}'}_{\le 1}) \ub{(g' - g' \wedge
  1)}_{\le |\nu' - 1|}
  + \ub{\Ebb (g' - \tilde{g}') (g' \wedge 1 - \tilde{g}')}_{= \text{(iii)}}
\]
So the first term is bounded above by $\Ebb(\nu' + 1)|\nu' + 1| = o(1)$ by LFC.

Notice that
\[
  \Ebb (g' - \tilde{g}') (g' \wedge 1 - \tilde{g}')
  = \Ebb g' \cdot g' \wedge 1
  + \Ebb \tilde{g}'^2
  - \Ebb g' \tilde{g}'
  - \Ebb \tilde{g}' \cdot g' \wedge 1
,\]
and each term is, up to $o(1)$, equal to $\Ebb \tilde{g}'^2$.
Indeed (let's do the hardest term),
\[
  \Ebb g' \cdot g' \wedge 1 - \Ebb \tilde{g}'^2
  = \Ebb_{x_1, x_2, x_3, x_4} \left[ g' \wedge 1 (x_{-1})
  \prod_{j = 2}^{4} g_{-j}(x_{-j}) - \tilde{g}'(x_{-1}) \prod_{j = 2}^4
  \tilde{g}_{-j} (x_{-j})  \right]
\]
but $\|g' \wedge 1 - \tilde{g}'\|_{\square} = o(1)$ by the claim, so done by
inductive hypothesis.
\qed