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We first derive the reflected attachment estimate for percolation. Put $R=[0,m]\times [0,2n]$, $m\ge n$, and $S=[0,n{]}^2$. Let $X(R)$ require an open vertical crossing of $S$ together with an open path in $R$ attaching it to the right side of $R$. ThenTo prove this, explore the leftmost vertical open crossing $P$ of $S$, revealing bonds on $P$ and to its left, but none strictly to its right. One can implement this exploration by following the boundary of the set accessible from the left using closed dual bonds; its right boundary is the extremal open crossing when that crossing exists. Thus fixing its outcome leaves right-side bonds in their original independent distribution.

For each possible explored $P$, reflect its geometry in $y=n$ and join it to its reflection to obtain a bottom-to-top separator $L$ of $R$. This reflected portion is only a geometrical separator and is not presumed open. Let $Y_{-}$ and $Y_{+}$ mean that an open path from the right boundary reaches, respectively, the lower or upper half of $L$, staying to its right until its endpoint. Any horizontal open crossing of $R$, followed from right to left until its first meeting with $L$, witnesses at least one of these events. Reflection in $y=n$ swaps $Y_{-}$ and $Y_{+}$ and preserves the independent measure. HenceThe bonds tested by $Y_{-}$ remain unexposed after specifying $P$; conditional on that exploration its probability is still the unconditional one for this deterministic separator. Its occurrence attaches the genuinely open lower crossing $P$ to the right side. Average over $P$ to obtain the attachment estimate.

Now take $R=[0,2n]\times [0,2n]$ and $R^{\prime }=[-n,n]\times [0,2n]$, with common corner square $S=[0,n{]}^2$. Use $X(R)$, its horizontal reflection in $R^{\prime }$, and a horizontal open crossing of $S$. The last crossing meets the vertical crossings of $S$ in both attachment events, so these three events give a horizontal crossing of $R\cup R^{\prime }$. All three events are increasing events, so the Harris lemma gives

For all aspect ratios, glue a width-$m_1$, height-$2s$ rectangle to a width-$m_2$, height-$2s$ rectangle with an overlap square of side $2s$. Their horizontal crossings and an overlap vertical crossing imply a crossing of their union. Again the Harris lemma givesAdding a $3s$-wide rectangle therefore yieldsFor height $n\ge 2$, let $s=\lfloor n/2\rfloor$, so $2s\le n$ and $s\ge n/3$. Choose an integer $k\ge \max (3,3\lambda )$. For $m\le \lambda n$, we have $m\le ks$; restricting to height $2s$ and extracting a shorter horizontal crossing givesFor $n=1$, the horizontal bottom-row path gives $h_{1/2}(m,1)\ge 2^{-\lceil \lambda \rceil }$ whenever the allowed integer $m\ge 1$ exists. Thus one may take $c_{\lambda }=\min (2^{17-8k},2^{-\lceil \lambda \rceil })>0$. This proves the requested uniform rectangle crossing in bond percolation bound, rather than assuming the Russo-Seymour-Welsh theorem.