[proofplan] We prove the relation by grouping all chains counted on the left according to the chain obtained after deleting the unique element whose rank lies strictly between $i$ and $k$. Fixing a chain with rank set $S$, the hypotheses identify two consecutive elements of the augmented chain, of ranks $i$ and $k$. The possible insertions of a rank-$j$ element are exactly the elements of rank $j$ in the open interval between them. The Eulerian alternating rank-sum relation on that interval gives the required signed contribution for the fixed chain, and summing over all chains with rank set $S$ gives the formula. [/proofplan] [step:Fix an $S$-chain and locate the rank gap from $i$ to $k$] Let $\mathcal C_S(P)$ denote the set of chains $C\subseteq P\setminus\{\hat{0},\hat{1}\}$ whose set of ranks is exactly $S$. Thus $|\mathcal C_S(P)|=f_S(P)$. Fix $C\in\mathcal C_S(P)$. Define the augmented chain $C^+:=C\cup\{\hat{0},\hat{1}\}$. Since $\rho(\hat{0})=0$, $\rho(\hat{1})=n+1$, and $\{i,k\}\subseteq S\cup\{0,n+1\}$, there are unique elements $x,y\in C^+$ such that $\rho(x)=i$ and $\rho(y)=k$. Because $i<k$, the chain order gives $x<y$. Since $S$ has no element in $\{i+1,\dots,k-1\}$, no element of $C^+$ has rank strictly between $i$ and $k$. Hence $x<y$ are consecutive elements of the augmented chain $C^+$. [/step] [step:Identify rank-$j$ insertions with elements in the open interval] For each integer $j$ with $i<j<k$, let $\mathcal E_j(C)$ be the set of chains $D\subseteq P\setminus\{\hat{0},\hat{1}\}$ whose rank set is $S\cup\{j\}$ and whose deletion of the unique rank-$j$ element is $C$. We claim that the map $\Phi_j:\mathcal E_j(C)\to \{z\in P:x<z<y,\ \rho(z)=j\}$, sending a chain $D$ to its unique element of rank $j$, is a bijection. Indeed, if $D\in\mathcal E_j(C)$ and $z=\Phi_j(D)$, then deleting $z$ leaves $C$. Since no rank of $C^+$ lies strictly between $i$ and $k$, the element $z$ must lie between the consecutive augmented-chain elements $x$ and $y$, so $x<z<y$ and $\rho(z)=j$. Conversely, if $z\in P$ satisfies $x<z<y$ and $\rho(z)=j$, then $C\cup\{z\}$ is a chain: every element of $C^+$ below $x$ is below $z$, every element of $C^+$ above $y$ is above $z$, and there is no element of $C^+$ strictly between $x$ and $y$. Its rank set is $S\cup\{j\}$, and deleting $z$ recovers $C$. Thus $\Phi_j$ is bijective. Therefore the contribution of the fixed chain $C$ to the coefficient of $f_{S\cup\{j\}}(P)$ equals $N_j(C):=\left|\{z\in P:x<z<y,\ \rho(z)=j\}\right|$. [/step] [step:Evaluate the signed insertion count by the Eulerian interval relation] Let $d:=k-i$ be the rank difference of the interval $[x,y]$. Since $i<k-1$, we have $d\ge 2$. By the explicit Eulerian poset alternating rank-sum hypothesis applied to the nontrivial interval $[x,y]$, we have $\sum_{z\in [x,y]}(-1)^{\rho(z)-\rho(x)}=0$. Separating the two endpoint terms from the interior terms gives $1+\sum_{j=i+1}^{k-1}(-1)^{j-i}N_j(C)+(-1)^{k-i}=0$. Hence $\sum_{j=i+1}^{k-1}(-1)^{j-i}N_j(C)=-1-(-1)^{k-i}$. Multiplying by $-1$ gives $\sum_{j=i+1}^{k-1}(-1)^{j-i-1}N_j(C)=1+(-1)^{k-i}$. Since $(-1)^{k-i}=-(-1)^{k-i-1}$, this is $\sum_{j=i+1}^{k-1}(-1)^{j-i-1}N_j(C)=1-(-1)^{k-i-1}$. [guided] We now compute the signed number of possible insertions into the fixed chain $C$. In this step, $C\in\mathcal C_S(P)$ is fixed, and $x,y\in C^+=C\cup\{\hat{0},\hat{1}\}$ are the unique elements with $\rho(x)=i$ and $\rho(y)=k$. The hypotheses on $S$, $i$, and $k$ imply that no rank of $C^+$ lies strictly between $i$ and $k$, so $x<y$ are consecutive elements of $C^+$. The interval involved is $[x,y]=\{z\in P:x\le z\le y\}$. Its rank difference is $d:=\rho(y)-\rho(x)=k-i$. The assumption $i<k-1$ means $d\ge 2$, so this is a nontrivial interval with possible interior ranks. The reason Eulerianity is relevant is that it converts a signed count of elements in an interval into a simple endpoint calculation. The formal statement includes the Eulerian poset alternating rank-sum condition for every nontrivial interval. Applied to $[x,y]$, with ranks measured relative to $\rho(x)=i$, this says $\sum_{z\in [x,y]}(-1)^{\rho(z)-i}=0$. We split this sum into three parts: the lower endpoint $x$, the upper endpoint $y$, and the interior elements $z$ with $x<z<y$. The lower endpoint contributes $1$, because $\rho(x)-i=0$. The upper endpoint contributes $(-1)^{k-i}$, because $\rho(y)=k$. For a fixed interior rank $j$ with $i<j<k$, the number of elements of rank $j$ in the open interval is $N_j(C)=\left|\{z\in P:x<z<y,\ \rho(z)=j\}\right|$. Therefore the Eulerian identity becomes $1+\sum_{j=i+1}^{k-1}(-1)^{j-i}N_j(C)+(-1)^{k-i}=0$. Solving for the interior alternating sum gives $\sum_{j=i+1}^{k-1}(-1)^{j-i}N_j(C)=-1-(-1)^{k-i}$. The Bayer-Billera relation uses the shifted sign $(-1)^{j-i-1}$, which is the negative of $(-1)^{j-i}$. Multiplying the previous identity by $-1$, we obtain $\sum_{j=i+1}^{k-1}(-1)^{j-i-1}N_j(C)=1+(-1)^{k-i}$. Finally, $(-1)^{k-i}=-(-1)^{k-i-1}$, so the fixed-chain contribution is $\sum_{j=i+1}^{k-1}(-1)^{j-i-1}N_j(C)=1-(-1)^{k-i-1}$. This is the desired local identity for one fixed chain $C$. [/guided] [/step] [step:Sum the fixed-chain identity over all chains of rank set $S$] Every chain counted by $f_{S\cup\{j\}}(P)$ has a unique rank-$j$ element, and deleting that element gives a unique chain in $\mathcal C_S(P)$, because $j\notin S$. Hence, for each $j$ with $i<j<k$, $f_{S\cup\{j\}}(P)=\sum_{C\in\mathcal C_S(P)}N_j(C)$. Using this decomposition and the fixed-chain identity from the previous step, we get $\sum_{j=i+1}^{k-1}(-1)^{j-i-1}f_{S\cup\{j\}}(P)=\sum_{C\in\mathcal C_S(P)}\sum_{j=i+1}^{k-1}(-1)^{j-i-1}N_j(C)$. For each $C\in\mathcal C_S(P)$, the inner sum equals $1-(-1)^{k-i-1}$, so $\sum_{j=i+1}^{k-1}(-1)^{j-i-1}f_{S\cup\{j\}}(P)=\sum_{C\in\mathcal C_S(P)}\left(1-(-1)^{k-i-1}\right)$. Since $|\mathcal C_S(P)|=f_S(P)$, this becomes $\sum_{j=i+1}^{k-1}(-1)^{j-i-1}f_{S\cup\{j\}}(P)=\left(1-(-1)^{k-i-1}\right)f_S(P)$. This is the claimed Bayer-Billera relation. [/step]