Ordered Composition — Examples

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.

Both compounds have type-total vector $(1,1)$, so [F1] applies. Their calibrated values are $\mathrm{\Sigma }(C)=1+1\mathrm{/}3=4\mathrm{/}3$ and $\mathrm{\Sigma }(D)=0+3=3$. Since $4\mathrm{/}3\le 3$, [F1] gives $C{⪯}_{\mathcal{Y}}D$. The theorem also cancels their common $2e_{{\zeta }_0}$, giving $C_0{⪯}_{\mathcal{Y}}{D}_0$. The value $3$ is permitted by [F2]; the representation theorem imposes no unit-interval bound on arbitrary states.

Reversing the comparison would require $3\le 4\mathrm{/}3$, which is false; hence this particular comparison is strict. No separate per-type numerical reading is treated as a proof of common-meter compatibility; that remains the stated hypothesis.

The multiplier $\lambda =(1,1)$ is nonnegative and satisfies ${\lambda }_1{\nu }_1+{\lambda }_2{\nu }_2=(1,-1)+(-1,1)=(0,0)$. But ${\lambda }_1{d}_1+{\lambda }_2{d}_2=-2+1=-1<0$, violating the necessary condition in [F1]; therefore no offset exists.

Directly, the two inequalities would require $b_1-b_2\ge 2$ and $-b_1+b_2\ge -1$, equivalently $b_1-b_2\le 1$. Adding the original inequalities gives $0\ge 1$, a contradiction. This is the positive-right-side zero-left certificate of [F2].

For $x=0$, $p(x)=0$. For $t>0$, $p(tx)=\max \{tx,0\}=t\max \{x,0\}=tp(x)$. Also $p(x+y)\le p(x)+p(y)$ because the right side is nonnegative and at least $x+y$. Every value is finite, so $p(x)<+\mathrm{\infty }$ iff $p(-x)<+\mathrm{\infty }$. Thus all assumptions in [F1] hold.

The map $L(x)=x\mathrm{/}2$ is real-linear. If $x\ge 0$, then $p(x)=x$, $p(-x)=0$, and $0\le x\mathrm{/}2\le x$. If $x\le 0$, then $p(x)=0$, $p(-x)=-x$, and $x\le x\mathrm{/}2\le 0$. These two cases include $x=0$ and give $-p(-x)\le L(x)\le p(x)$ for all $x$.