Catalyst-chain cost characterizes cross-type transitions

If $x\prec y$, the one-edge witness with an identity catalyst has cost ${\sigma }_{{\mathrm{\Gamma }}^{\mathrm{\prime }}}(y)-{\sigma }_{\mathrm{\Gamma }}(x)$. Hence $F(\mathrm{\Gamma },{\mathrm{\Gamma }}^{\mathrm{\prime }})\le {\sigma }_{{\mathrm{\Gamma }}^{\mathrm{\prime }}}(y)-{\sigma }_{\mathrm{\Gamma }}(x)$, which is the forward inequality, including the case $F=+\mathrm{\infty }$ where no such transition can exist.

Conversely assume the displayed entropy inequality. It is impossible when $F=+\mathrm{\infty }$. Fix an arbitrary natural index $n\in \mathbb{N}$, use ${ϵ}_n$ from [F5] and choose a decorated witness of real cost $c_n\le \mathrm{\Delta }\sigma +h{ϵ}_n$, where $\mathrm{\Delta }\sigma ={\sigma }_{{\mathrm{\Gamma }}^{\mathrm{\prime }}}(y)-{\sigma }_{\mathrm{\Gamma }}(x)$ and $h=\sigma (z_1)-\sigma (z_0)>0$ is the entropy gap of a fixed strict meter pair from [F1]. Such witnesses exist from the definition of infimum, also if $F=-\mathrm{\infty }$. Write their types as $T_0=\mathrm{\Gamma }\times C,T_1,\dots ,T_N={\mathrm{\Gamma }}^{\mathrm{\prime }}\times C$, their edges as $a_k\prec b_{k+1}$, and their endpoints as $a_0=(\tilde{x},c_0)$, $b_N=(\tilde{y},d_0)$.

Put ${\delta }_n={ϵ}_n$. In one finite compound type form $L_n=(x,b_N,b_1,\dots ,b_{N-1},{\delta }_n{z}_0),R_n=(a_0,y,a_1,\dots ,a_{N-1},{\delta }_n{z}_1)\mathrm{.}$ These have the same composition: both contain $\mathrm{\Gamma }$, ${\mathrm{\Gamma }}^{\mathrm{\prime }}$, the catalyst $C$, every intermediate type $T_1,\dots ,T_{N-1}$, and one scaled meter state, up to reassociation and permutation. Additivity of $\sigma$ and the definition of chain cost give $\sigma (R_n)-\sigma (L_n)=\mathrm{\Delta }\sigma -c_n+{\delta }_{n}h\ge 0.$ Thus the same-composition entropy representation [F1] gives $L_n\prec R_n$.

Starting from $R_n$, use consistency on $a_0\prec b_1$ with $y,{\delta }_n{z}_1,a_1,\dots ,a_{N-1}$ as spectators. The result contains $b_1$ in place of $a_0$. Next use $a_1\prec b_2$ with $y,{\delta }_n{z}_1,b_1,a_2,\dots ,a_{N-1}$ as spectators, and continue in order. At stage $k$, the current compound contains $a_k$, the earlier outputs $b_1,\dots ,b_k$, all later inputs $a_{k+1},\dots ,a_{N-1}$, and the unchanged $y,{\delta }_n{z}_1$. After the last edge, transitivity gives $R_n\prec R_n^{\mathrm{\prime }}$, where $R_n^{\mathrm{\prime }}=(b_1,\dots ,b_N,y,{\delta }_n{z}_1)$. Compose this with $L_n\prec R_n$ from step 2.1. The compounds $L_n$ and $R_n^{\mathrm{\prime }}$ share the finite spectator $b_1+\cdots +b_N$; cancel it by [F4]. The result is $x+{\delta }_n{z}_0\prec y+{\delta }_n{z}_1$ for this $n$. Since $n$ was arbitrary, this comparison holds for every natural index; the chain witnesses need not be chosen simultaneously.

By [F5], ${\delta }_n\downarrow 0$, stability in [A1] with bases $x,y$ and fixed positive spectators $z_0,z_1$ yields $x\prec y$. This proves the reverse implication without assuming that any infimum is attained.