Spectator cancellation without comparison completeness

Suppose $A+C\prec B+C$ 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 $A\mathrm{/}r+C\mathrm{/}r\prec B\mathrm{/}r+C\mathrm{/}r$. 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, $A+{ϵ}_{k}C\prec B+{ϵ}_{k}C$. 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 $A\prec B$. If $m(C)=0$, nonnegativity implies $C=0$ and the conclusion is the premise itself. Conversely, $A\prec B$ gives $A+C\prec B+C$ by consistency.

For equal-total signed maps $u,v$, the difference $d=v-u$ has finite support and total coefficient zero, so $m(d^{+})=m(d^{-})$. If $u,v$ are nonnegative, their pointwise common part $k(x)=\min \{u(x),v(x)\}$ gives $u=k+d^{-}$ and $v=k+d^{+}$. Cancellation in step 1.1 proves $u\prec v$ exactly when $d^{-}\prec d^{+}$, which by [F2] is $u{⪯}_{s}v$; adding a common signed map leaves $d$ unchanged by [F2].

If $u{⪯}_{s}v$ and $v{⪯}_{s}w$, [F2] identifies these with comparisons of $(v-u{)}^{-}$ to $(v-u{)}^{+}$ and $(w-v{)}^{-}$ to $(w-v{)}^{+}$. Consistency combines them. Their pointwise common minimum cancels by step 1.1, leaving $(w-u{)}^{-}\prec (w-u{)}^{+}$, which [F2] identifies as $u{⪯}_{s}w$. The same finite-sum argument proves closure under addition; reflexivity follows from reflexivity of the zero compound.