The proof consists of two parts. First we need to show that a vector $\vct{b}$ satisfying the linear system exists and further can be computed in $O(m)$ time. Second we need to show that $\numocc{G}{\tri}, \numocc{G}{\threedis}$ can indeed be computed in time $O(1)$.
Note that the LHS of \Cref{eq:b1-alg-2} is obtained using \cref{eq:2pd-3d} and \cref{eq:3p-3tri} and is indeed the product $\vct{M}[1]\cdot\vct{x}[1]$. Further note that this product is equal to the RHS of \Cref{eq:b1-alg-2}, where every term is computable in $O(m)$ time (by equations (\ref{eq:1e})-(\ref{eq:3p-3tri})). We set $\vct{b}[1]$ to the RHS of \Cref{eq:b1-alg-2}.
We follow the same process in deriving an equality for $G^{(2)}$. Replacing occurrences of $G$ with $G^{(2)}$, we obtain an equation (below) of the form of \cref{eq:b1-alg-2} for $G^{(2)}$. Substituting identities from \cref{lem:3m-G2} and \Cref{lem:tri} we obtain
The steps to obtaining \cref{eq:b2-final} are analogous to the derivation immediately preceding. As in the previous derivation, note that the LHS of \Cref{eq:b2-final} is the same as $\vct{M}[2]\cdot\vct{x}[2]$. The RHS of \Cref{eq:b2-final} has terms all computable (by equations (\ref{eq:1e})-(\ref{eq:3p-3tri})) in $O(m)$ time. Setting $\vct{b}[2]$ to the RHS then completes the proof of step 1.
From \Cref{eq:det-final} it can easily be seen that the roots of $\dtrm{\vct{M}}$ are $0, 1,$ and $3$. Hence there are no roots in $(0, 1)$ and \Cref{lem:lin-sys} follows.
For the sake of contradiction, assume that for any $G$, we can compute $\rpoly_{G}^3(\prob,\dots,\prob)$ in $o\inparen{m^{1+\eps_0}}$ time.
Let $G$ be the input graph. It is easy to see that one can compute the expression tree for $\poly_{G}^3(\vct{X})$ in $O(m)$ time. Then by \Cref{th:single-p} we can compute $\numocc{G}{\tri}$ in further time $o\inparen{m^{1+\eps_0}}+O(m)$. Thus, the overall, reduction takes $o\inparen{m^{1+\eps_0}}+O(m)= o\inparen{m^{1+\eps_0}}$ time, which violates \Cref{conj:graph}.