Ordered Composition

Let $E$ be any finite subset of $X$ containing $\mathrm{supp}(c)$. On $E\setminus \mathrm{supp}(c)$ the function is zero, so the finite sum over $E$ equals the finite sum over $\mathrm{supp}(c)$ by [F2]. If $E$ and $E^{\mathrm{\prime }}$ both contain the support, compare each sum to the sum over the finite union $E\cup E^{\mathrm{\prime }}$; hence the value is independent of the chosen finite summation set.

The support of $c+d$ is contained in the finite union of the supports of $c$ and $d$. Sum over that union; distributivity in [F2] gives $m(c+d)=m(c)+m(d)$, since zero terms outside each individual support contribute nothing by step 1.1. The support of $tc$ equals that of $c$ for $t>0$, and scalar linearity of the finite sum gives $m(tc)=tm(c)$. For $t=0$, $tc=0$ and both $m(tc)$ and $tm(c)$ are zero. The empty sum is zero, so $m(0)=0$.

The support of $e_x$ is the singleton $\{x\}$, whose one summand is $1$, so the recursion in [F2] gives $m(e_x)=0+1=1$. If $m(c)=0$, sum its nonnegative coefficients over its finite support. Clause 4 in [F2] forces every summand to be zero; coefficients off the support are also zero, hence $c=0$. Conversely $c=0$ has zero mass by step 2.1. Thus all the claimed mass identities hold, including the empty and unit-state cases.

The set $N_{\mathbb{R}}:=\{{\iota }_{\mathbb{N}}(n):n\in \mathbb{N}\}$ is nonempty. If it were bounded above, completeness [F1] would give $s=\sup N_{\mathbb{R}}$. Since $1>0$ by [F4], $s-1<s$, so $s-1$ is not an upper bound; choose $n\in \mathbb{N}$ with ${\iota }_{\mathbb{N}}(n)>s-1$. By the successor rule in [F3], ${\iota }_{\mathbb{N}}(\sigma (n))={\iota }_{\mathbb{N}}(n)+1>s$, while $\sigma (n)\in \mathbb{N}$, contradicting that $s$ is an upper bound. Hence $N_{\mathbb{R}}$ is unbounded above.

For every $n\in \mathbb{N}$, $\sigma (n)=1+n$ by natural-addition commutativity and the successor rule, so $1\le \sigma (n)$ by the definition of natural order; hence ${\iota }_{\mathbb{N}}(\sigma (n))>0$ by [F3]. The successor recursion gives ${\iota }_{\mathbb{N}}(\sigma (\sigma (n)))={\iota }_{\mathbb{N}}(\sigma (n))+1>{\iota }_{\mathbb{N}}(\sigma (n))$, since $1>0$. Reciprocation reverses this strict inequality by [F4], hence $0<{ϵ}_n$ and ${ϵ}_{n+1}<{ϵ}_n$. Now fix rational $q>0$ and let $\widehat{q}>0$ be its real image. By step 1.1 choose $N\in \mathbb{N}$ with ${\iota }_{\mathbb{N}}(\sigma (N))>1\mathrm{/}\widehat{q}$. If $n\ge N$, then $\sigma (N)\le \sigma (n)$ by [F2] and Order is compatible with addition, and the increasing-map clause of [F3] gives ${\iota }_{\mathbb{N}}(\sigma (n))\ge {\iota }_{\mathbb{N}}(\sigma (N))$. Therefore $0<{ϵ}_n\le 1\mathrm{/}{\iota }_{\mathbb{N}}(\sigma (N))<\widehat{q}$. This is convergence to zero by [F5].

For equal-total $u,v$, $d=v-u$ has finite support and total sum zero. Pointwise $d=d^{+}-d^{-}$, so finite-sum linearity gives $m(d^{+})-m(d^{-})=0$; both maps are finite-support and nonnegative. Thus the definition in [F2] compares equal-mass compounds. If $u,v$ are nonnegative, set $k(x)=\min \{u(x),v(x)\}$; then $u=k+(v-u{)}^{-}$ and $v=k+(v-u{)}^{+}$.

Suppose $A+C⪯B+C$ and $m(C)>0$. For each $k\in \mathbb{N}$, set $n=\sigma (k)$ and $r={\iota }_{\mathbb{N}}(n)>0$, and write $A\mathrm{/}r=(1\mathrm{/}r)A$. The embedding's successor rule and finite induction show that the sum of $n$ copies of $A\mathrm{/}r$ is $r(A\mathrm{/}r)=A$, and likewise for $B$. Scaling the given comparison by $1\mathrm{/}r$ gives $A\mathrm{/}r+C\mathrm{/}r⪯B\mathrm{/}r+C\mathrm{/}r$. By composition consistency, replace the $n$ copies of $A\mathrm{/}r$ one at a time by $B\mathrm{/}r$, keeping the already replaced copies, the remaining copies, and $C\mathrm{/}r$ as spectators; after these finitely many replacements, $A+{ϵ}_{k}C⪯B+{ϵ}_{k}C$. This holds for every natural index $k$. By [F3], ${ϵ}_k>0$ and ${ϵ}_k\to 0$, so stability with $Z_0=Z_1=C$ gives $A⪯B$. If $m(C)=0$, nonnegativity implies $C=0$ and the conclusion is the premise itself. Conversely, $A⪯B$ gives $A+C⪯B+C$ by consistency.

For nonnegative $u,v$, step 1.1 gives $u=k+(v-u{)}^{-}$ and $v=k+(v-u{)}^{+}$. Cancellation from step 1.2 shows $u⪯v$ exactly when $(v-u{)}^{-}⪯(v-u{)}^{+}$, which is $u{⪯}_{s}v$. For general signed maps, adding a common map leaves $v-u$ unchanged, so it leaves signed comparison unchanged.

If $u{⪯}_{s}v$ and $v{⪯}_{s}w$, consistency gives $N⪯P$, where $N:=(v-u{)}^{-}+(w-v{)}^{-}$ and $P:=(v-u{)}^{+}+(w-v{)}^{+}$. Pointwise $P-N=w-u$. Set $K(x):=\min \{P(x),N(x)\}$. If $P(x)\ge N(x)$ then $P(x)-K(x)=(w-u{)}^{+}(x)$ and $N(x)-K(x)=(w-u{)}^{-}(x)$; if $P(x)<N(x)$ the same identities follow with the two sides exchanged. Thus $P=K+(w-u{)}^{+}$ and $N=K+(w-u{)}^{-}$, with all four maps finite-support and nonnegative. Cancellation from step 1.2 gives $(w-u{)}^{-}⪯(w-u{)}^{+}$, hence $u{⪯}_{s}w$. For two independent signed comparisons, the same pointwise decomposition of the sums of their positive and negative parts proves closure under adding comparisons. Reflexivity follows because $0⪯0$.

Put $\delta :=\mu -\lambda$. The coefficient difference is $R_{\mu }-R_{\lambda }=\delta (e_{x_1}-e_{x_0})$ and has total zero. If $\delta >0$, its negative and positive parts are $\delta e_{x_0}$ and $\delta e_{x_1}$; if $\delta <0$, they are $(-\delta )e_{x_1}$ and $(-\delta )e_{x_0}$; if $\delta =0$, both are zero.

If $\delta =0$, reflexivity gives $R_{\lambda }{⪯}_s{R}_{\mu }$. If $\delta >0$, [F1] reduces this comparison to $\delta e_{x_0}⪯\delta e_{x_1}$, which follows by positive scaling of $x_0⪯x_1$ by $\delta$. Hence $\lambda <\mu$ implies $R_{\lambda }{⪯}_s{R}_{\mu }$.

If $\delta <0$ and $R_{\lambda }{⪯}_s{R}_{\mu }$, [F1] gives $c{e}_{x_1}⪯c{e}_{x_0}$ for $c:=-\delta >0$. In the ordered real field, $c^{-1}>0$: $c^{-1}\mathrm{\ne }0$, and if $-c^{-1}>0$ then closure of the positive cone under multiplication gives $c(-c^{-1})=-1>0$. But $1>0$: since $1\mathrm{\ne }0$, trichotomy gives either $1>0$ or $-1>0$; the latter would, by positive-product closure, also give $(-1{)}^2=1>0$, contradicting trichotomy. Thus $-1>0$ is impossible, and so $c^{-1}>0$. Positive scaling by $c^{-1}$ therefore implies $x_1⪯x_0$, contrary to [A1]. Thus this comparison fails when $\lambda >\mu$. The three cases prove the equivalence.

Write $q_n:={\iota }_{\mathbb{N}}(\sigma (n))>0$, so ${ϵ}_n=1\mathrm{/}{q}_n$ by [F4]. Suppose no $n\in \mathbb{N}$ satisfies $(q_n+1)e_{x_0}⪯q_n{e}_{x_1}+e_x$. Both sides have mass $q_n+1$, so CH gives $q_n{e}_{x_1}+e_x⪯(q_n+1)e_{x_0}$. Scale by $1\mathrm{/}{q}_n={ϵ}_n$ to obtain $e_{x_1}+{ϵ}_n{e}_x⪯e_{x_0}+{ϵ}_n{e}_{x_0}$. Apply stability to the sequence ${ϵ}_n$, with bases $e_{x_1},e_{x_0}$ and positive unit-mass spectators $e_x,e_{x_0}$. It yields $x_1⪯x_0$, contradicting strictness. Thus some $n$ has $(q_n+1)e_{x_0}⪯q_n{e}_{x_1}+e_x$, which by signed translation and cancellation in [F2] is $R_{-q_n}{⪯}_s{e}_x$; hence $-q_n\in L_x$.

Suppose $L_x$ is unbounded above. For each $n\in \mathbb{N}$, unboundedness supplies $\lambda \in L_x$ with $\lambda >q_n$; [F3] gives $R_{q_n}{⪯}_s{R}_{\lambda }$, and transitivity gives $q_n\in L_x$. Since $q_n\ge 1$ by [F4], adding the common compound $q_n{e}_{x_0}$ to $R_{q_n}$ and $e_x$ makes them $e_{x_0}+q_n{e}_{x_1}$ and $e_x+q_n{e}_{x_0}$. Thus [F2] identifies $R_{q_n}{⪯}_s{e}_x$ with the compound comparison between those maps. Scaling by ${ϵ}_n=1\mathrm{/}{q}_n$ gives $e_{x_1}+{ϵ}_n{e}_{x_0}⪯e_{x_0}+{ϵ}_n{e}_x$. Stability with bases $e_{x_1},e_{x_0}$ and positive unit-mass spectators $e_{x_0},e_x$ now yields $x_1⪯x_0$, again a contradiction. Therefore $L_x$ is bounded above.

If $\lambda \in L_x$ and $\mu \le \lambda$, [F3] gives $R_{\mu }{⪯}_s{R}_{\lambda }$; signed transitivity and $R_{\lambda }{⪯}_s{e}_x$ imply $\mu \in L_x$. Thus $L_x$ is downward closed. By steps 1.1 and 1.2 it is nonempty and bounded above, so [F5] supplies the real number $s:=\sup L_x$.

For each $n\in \mathbb{N}$, $s-{ϵ}_n$ is not an upper bound of $L_x$, so some $\lambda \in L_x$ has $\lambda >s-{ϵ}_n$. Downward closure gives $s-{ϵ}_n\in L_x$. Consistency then gives $R_s+{ϵ}_n{e}_{x_0}{⪯}_s{e}_x+{ϵ}_n{e}_{x_1}$, because $R_{s-{ϵ}_n}+{ϵ}_n{e}_{x_1}=R_s+{ϵ}_n{e}_{x_0}$. Let $K:=R_s^{-}$. Adding the fixed nonnegative map $K$ to both sides converts this signed comparison into the compound comparison $R_s^{+}+{ϵ}_n{e}_{x_0}⪯e_x+K+{ϵ}_n{e}_{x_1}$. Their bases have equal mass because $m(R_s^{+})-m(K)=1=m(e_x)$. Stability with positive unit-mass spectators $e_{x_0},e_{x_1}$ yields $R_s^{+}⪯e_x+K$, that is, $R_s{⪯}_s{e}_x$.

Since $s$ is an upper bound, $s+{ϵ}_n\mathrm{\notin }{L}_x$ for every $n\in \mathbb{N}$. To derive the reverse signed comparison from CH, set $H_n:=R_{s+{ϵ}_n}^{-}$. The translated maps $e_x+H_n$ and $R_{s+{ϵ}_n}+H_n=R_{s+{ϵ}_n}^{+}$ are nonnegative compounds. Since $R_{s+{ϵ}_n}=R_{s+{ϵ}_n}^{+}-H_n$ has total coefficient one, finite-sum linearity gives $m(R_{s+{ϵ}_n}^{+})-m(H_n)=1$; with $m(e_x)=1$, the translated compounds have equal mass $1+m(H_n)$. CH applies to this pair. If it gave $R_{s+{ϵ}_n}^{+}⪯e_x+H_n$, signed translation invariance in [F2] would imply $R_{s+{ϵ}_n}{⪯}_s{e}_x$, contrary to $s+{ϵ}_n\mathrm{\notin }{L}_x$. Therefore it gives $e_x+H_n⪯R_{s+{ϵ}_n}^{+}$, which by [F2] is $e_x{⪯}_s{R}_{s+{ϵ}_n}$. Add ${ϵ}_n{e}_{x_0}$ to both signed maps and use $R_{s+{ϵ}_n}+{ϵ}_n{e}_{x_0}=R_s+{ϵ}_n{e}_{x_1}$. Then add $K=R_s^{-}$ to both; the resulting equal-mass compound comparison is $e_x+K+{ϵ}_n{e}_{x_0}⪯R_s^{+}+{ϵ}_n{e}_{x_1}$. Stability with bases $e_x+K,R_s^{+}$ and positive unit-mass spectators $e_{x_0},e_{x_1}$ gives $e_x+K⪯R_s^{+}$, or $e_x{⪯}_s{R}_s$. Together with step 3.1 this proves $R_s{\sim }_s{e}_x$ and the claimed attainment.

In the strict-reference case, [F3] gives $R_{\lambda }{⪯}_s{e}_{x_0}$ exactly for $\lambda \le 0$, and $R_{\lambda }{⪯}_s{e}_{x_1}$ exactly for $\lambda \le 1$. Their suprema are therefore $S(x_0)=0$ and $S(x_1)=1$.

Let $C={\sum }_i{a}_i{e}_{x_i}$ have mass $M$ and set $q_C:={\sum }_i{a}_{i}S(x_i)$. Discard any zero terms in this presentation using [F4]. If the support is empty, $C=0$, $M=q_C=0$ and $C_R=0$, so signed reflexivity gives the claimed equivalence. Otherwise each remaining coefficient is strictly positive. For each remaining constituent, [F2] gives $e_{x_i}{\sim }_s{R}_{S(x_i)}$; positive scaling and finite repeated use of composition combine these into $C{\sim }_s{C}_R$, where $C_R:={\sum }_i{a}_i{R}_{S(x_i)}=(M-q_C)e_{x_0}+q_C{e}_{x_1}$. This calculation is valid for signed coefficients because the signed relation is additive and translation-invariant by [F1].

Apply step 1.2 to both $A$ and $B$, and write $q_A,q_B$ for the resulting scalar sums. Since $A{\sim }_s{A}_R$ and $B{\sim }_s{B}_R$, transitivity gives $A⪯B$ exactly when $A_R{⪯}_s{B}_R$. Their masses agree, and $B_R-A_R=(q_B-q_A)(e_{x_1}-e_{x_0})$; [F3] therefore gives $A_R{⪯}_s{B}_R$ exactly when $q_A\le q_B$. This proves the representation formula.

In the no-strict-pair case, take any $x,y\in X$. By [F4] their unit compounds have equal mass one, so CH gives one of $x⪯y$ or $y⪯x$; if exactly one held it would be a strict pair, so both hold. Fix $r\in X$. Let $C$ have mass $M$. If $M=0$, [F4] gives $C=0$, and reflexivity handles this empty compound without applying the positive-scaling axiom at zero. If $M>0$, use the canonical finite decomposition over the support from [F4]; every coefficient there is positive. Scale each constituent equivalence $e_x\sim e_r$ by its positive coefficient and compose the finitely many comparisons to obtain $C\sim M{e}_r$. Thus equal-mass $A,B$ are both empty when their common mass is zero, and otherwise each is equivalent to the same $M{e}_r$; hence they are equivalent in both cases. For a constant $S(x)=c$, both compound values equal $cM$, including zero for empty compounds, so the representation equivalence holds.

Fix $x\in X$ and put $s=S(x)$. The equivalence $x{\sim }_s{R}_s$ in [F1] says that the positive and negative parts of their coefficient difference compare in both directions. Since $T$ represents the same order, their finite $T$-sums are equal. By its definition on finite compounds and linearity of finite sums, the signed value of $R_s$ is $(1-s)T(x_0)+sT(x_1)$, including when $s\mathrm{\notin }[0,1]$. Thus $T(x)=T(x_0)+[T(x_1)-T(x_0)]S(x)$.

Because $x_0⪯x_1$ but not conversely, the exact representation property of $T$ gives $T(x_0)<T(x_1)$. Hence $a:=T(x_1)-T(x_0)>0$ and $b:=T(x_0)$ give the asserted form. Conversely, multiplying both sides of any represented-value inequality by $a>0$ and adding the same $b$ times the common compound mass preserves and reflects its order; therefore $aS+b$ represents the same comparisons. The parameters are unique by evaluating at $x_0$ and $x_1$, where $S$ equals $0$ and $1$.

For each constituent $(i,x_{ik},z_0)$ of $C$, its mass-two shared compound $u_{i,x_{ik}}$ is equivalent under ${\sim }_{s,\mathcal{Y}}$ by [A1] to $e_{{\rho }_i}+(1-{\sigma }_i(x_{ik}))e_{{\zeta }_0}+{\sigma }_i(x_{ik})e_{{\zeta }_1}$. Each side has total coefficient two, so the signed substitution is defined. Scale by $t_{ik}>0$ and combine the finitely many comparisons using the comparison axioms and [F2]. The resulting reference-system part is ${\sum }_i{T}_i{e}_{{\rho }_i}$, independent of the chosen states. The meter part has coefficients $T-\mathrm{\Sigma }(C)$ on $e_{{\zeta }_0}$ and $\mathrm{\Sigma }(C)$ on $e_{{\zeta }_1}$, where $T:={\sum }_i{T}_i$. Hence $C$ is equivalent under ${\sim }_{s,\mathcal{Y}}$ to that fixed reference-system compound plus $(T-\mathrm{\Sigma }(C))e_{{\zeta }_0}+\mathrm{\Sigma }(C)e_{{\zeta }_1}$, whose total coefficient is $T+T=2T$. If the collection is empty, both expressions are zero and the equivalence is reflexivity, requiring no order scaling at zero.