\begin{lemma}
\label{l:expon}
Let $d=2$ and $\lambda > \lambda_{c}$, then there exist constants $a_{1}$ and $c$ such that $Pr\left[LR_{a}\right] \geq 1-e^{-ca}$ for all $a \geq a_{1}$.
\end{lemma}
\begin{proof}
Let $\delta < 1$ be some real number. Consider the box $H_{i}=\left[ib,\left(i+2\right)b\right] \times \left[0,b\right]$ for $i=0,1,2$. We choose $b>0$ such that $Pr\left[LR_{b}\right] \geq 1 - \frac{\delta}{25}$ and $Pr\left[SLR_{b}\right] \geq 1 - \frac{\delta}{25}$ for $\delta$ (we know that such $b$ exists from the previous lemma). Let $V_{1}=H_{0} \cap H_{1}$ and $V_{2}=H_{1} \cap H_{2}$.

The occurrence of a vertical crossing in $V_{1}$ and $V_{2}$ together with horizontal crossings in $H_{0}$, $H_{1}$ and $H_{2}$ imply the event $LLR_{b}$. See Figure~\ref{fig:LLRB}. Thus the probability that $LLR_{b}$ does not happen is, by Boole's inequality, $$1-Pr\left[LLR_{b}\right] \leq \frac{5\delta}{25}=\frac{\delta}{5}.$$

\begin{figure}
	\centering
	\includegraphics{figLLRB.pdf}
	\caption{Occurence of $LLR_{b}$}
	\label{fig:LLRB}
\end{figure}

Since $B\left(2b,2\right)$ is a disjoint union of two copies of $B\left(b,4\right)$, by independence we get $$1-Pr\left[LR_{2b}\right] \leq \left(1-Pr\left[LLR_{b}\right]\right)^{2}\leq \frac{\delta^{2}}{25}$$ and similarly for $B\left(2b,1\right)$ and $B\left(b,2\right)$ we have $$1-Pr\left[SLR_{2b}\right] \leq \left(\frac{\delta}{25}\right)^{2} \leq \frac{\delta^{2}}{25}.$$

Using induction we obtain $$1-Pr\left[LR_{2^{k}b}\right]\leq \frac{\delta^{2^{k}}}{25}$$ and $$1-Pr\left[SLR_{2^{k}b}\right]\leq \frac{\delta^{2^{k}}}{25}$$ for every positive integer $k$. By the previous lemma there exists $a_{0}$ such that $Pr\left[LR_{a}\right] \geq \frac{49}{50}$ and $Pr\left[SLR_{a}\right] \geq \frac{49}{50}$ for every $a\geq a_{0}$. For $a>a_{0}$ we choose $\delta=\frac{1}{2}$ and an integer $m$ such that $2^{m}a_{0}\leq a < 2^{m+1}a_{0}$. Afterwards we have $$1-Pr\left[LR_{2^{m}b}\right]  \leq \frac{2^{-2^{m}}}{25} < \frac{2^{\frac{-a}{2a_{0}}}}{25}=e^{-ca}$$ and $$1-Pr\left[SLR_{2^{m}b}\right] \leq \frac{2^{-2^{m}}}{25}< \frac{2^{\frac{-a}{2a_{0}}}}{25}=e^{-ca}.$$
\end{proof}