Two-state reference order

Reflexivity and transitivity follow from those properties of $\le$ on the $b$ coefficient, and CH follows because $\le$ is total. Adding a common compound adds the same $b$ coefficient to each side, so consistency holds. Multiplying both coefficients by $t>0$ preserves the inequality. Splitting a state coefficient into two positive parts preserves the total coefficient of $b$, so splitting/recombination gives equivalence. Finally $e_a⪯e_b$ because $0\le 1$, while $e_b⋠e_a$ because $1\nleqq 0$.

For stability, suppose $A+{ϵ}_n{Z}_0⪯B+{ϵ}_n{Z}_1$ for every $n$, with ${ϵ}_n\to 0$ and fixed positive $Z_0,Z_1$. If $A(b)>B(b)$, let $\delta =A(b)-B(b)>0$ and $q:=\mathrm{\mid }{Z}_0(b)-Z_1(b)\mathrm{\mid }$. If $q=0$, every perturbed difference remains $\delta >0$; if $q>0$, choose $n$ with ${ϵ}_{n}q<\delta$. In either case some perturbed $b$-coefficient difference is positive, contradicting the assumed comparison. Thus $A(b)\le B(b)$, which is $A⪯B$.

The reference pair $x_0=a,x_1=b$ is strict by step 1.1. Therefore [F2] applies and yields $R_{\lambda }{⪯}_s{R}_{\mu }$ iff $\lambda \le \mu$ for all real $\lambda ,\mu$, including parameters outside $[0,1]$. The signed parameters are handled by the signed comparison definition, not by treating $R_{\lambda }$ as a nonnegative compound.