 the difference in the method of bounding diagrams for the expansion coefficients. Take the $0^\text{th}$-expansion coefficient for example. For percolation, the BK inequality simply tells us that $$\begin{aligned}
{\label{eq:pi0perc-comp}}
\pi_p^{{\scriptscriptstyle}(0)}(x)\leq{{\mathbb P}}_p(o{\underset{\raisebox{5pt}{${\scriptscriptstyle}$}}
 {\overset{}{\longleftrightarrow}}}x)^2.\end{aligned}$$ For the ferromagnetic Ising model, on the other hand, we first recall [(\[eq:pi0-def\])]{}, i.e., $$\begin{aligned}
\pi_\Lambda^{{\scriptscriptstyle}(0)}(x)=\sum_{{\partial}{{\bf n}}=o{\vartriangle}x}\frac{w_\Lambda({{\bf n}})}
 {Z_\Lambda}\,{\mathbbm{1}{\raisebox{-2pt}{$\scriptstyle \{o{\underset{\raisebox{5pt}{${\scriptscriptstyle}{{\bf n}}$}}
 {\overset{}{\Longleftrightarrow}}}x\}$}}},\end{aligned}$$ where $w_\Lambda({{\bf n}})/Z_\Lambda\ge0$. Due to the indicator, every current configuration ${{\bf n}}\in{{\mathbb Z}_+}^{{{\mathbb B}}_\Lambda}$ that gives nonzero contribution has at least *two bond-disjoint* paths $\zeta_1,\zeta_2$ from $o$ to $x$ such that $n_b>0$ for all $b\in\zeta_1{\:\Dot{\cup}\:}\zeta_2$. Also, due to the source constraint, there should be at least one path $\zeta$ from $o$ to $x$ such that $n_b$ is odd for all $b\in\zeta$. Suppose, for example, that $\zeta=\zeta_1$ and that $n_b$ for $b\in\zeta_2$ are all positive-even. Since a positive-even integer can split into two odd integers, on the labeled graph ${{\mathbb G}}_{{\bf n}}$ with ${\partial}{{\mathbb G}}_{{\bf n}}=o{\vartriangle}x$ (recall the notation introduced above [(\[eq:Sbefore\])]{}) there are at least *three edge-disjoint* paths from $o$ to $x$. This observation naturally leads us to expect that $$\begin{aligned}
{\label{eq:pi0-comp}}
\pi_\Lambda^{{\scriptscriptstyle}(0)}(x)\leq{{\langle \varphi_o\varphi_x \rangle}}_\Lambda^3\end{aligned}$$ holds for the ferromagnetic Ising model. This naive argument to justify [(\[eq:pi0-comp\])]{} will be made rigorous in Section \[s:bounds\] by taking account of partition functions.

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