First, we argue that $\sim$ is an equivalence relation. The reflectiveness and symmetry of the relation follows immediately from the definition. Regarding the transitivity, we can use the axioms, in particular, S2, to see that if $b=\mathbb{S}^{n}a$ then $c=\mathbb{S}^{n+k}a$ iff $c=\mathbb{S}^{k}b$ , and $b=\mathbb{S}^{n+k}c$ iff $a=\mathbb{S}^{k}c$ , from which the transitivity of $\sim$ follows.

Now, we show that $\prec$ is well-defined. Suppose that $[a]\prec[b]$ . We need to show that if $c\sim a$ then $[c]\prec[b]$ , and if $c\sim b$ , then $[a]\prec[c]$ . If $c\sim a$ , then $c\nsim b$ ($\sim$ is an equivalence relation), and either $c<^{\mathfrak{A}}a$ or $c=a$ , in which case $c<^{\mathfrak{A}}b$ , or $c=\mathbb{S}^{n}a$ , in which case if we assume $b=c$ or $a<^{\mathfrak{A}}b<^{\mathfrak{A}}c$ , then $b=\mathbb{S}^{k}a$ for some $0<k\le n$ (by S2), contradicting the assumption that $a\nsim b$ . In either case, $c\nsim b$ and $c<^{\mathfrak{A}}b$ , implying $[c]\prec[b]$ . Similarly, we consider the case $c\sim b$ to show $[a]\prec[c]$ .

Finally, we ensure that the ordering properties hold for $\prec$ .

1. Trichotomy. For every $a$ and $b$ , either $a\sim b$ ($[a]=[b]$ ) or $a\nsim b$ , hence, $a\neq b$ , and either $a<^{\mathfrak{A}}b$ ($[a]\prec[b]$ ) or $b<^{\mathfrak{A}}a$ ($[b]\prec[a]$ ).
2. Antisymmetry. For every $a$ and $b$ , if $a\nsim b$ and $a<^{\mathfrak{A}}b$ ($[a]\prec[b]$ ) then not $b<^{\mathfrak{A}}a$ ($[b]\nprec[a]$ ). Alternatively, $a\sim b$ or $a\nless^{\mathfrak{A}}b$ ($[a]\nprec[b]$ ) iff $b\sim a$ ($[b]=[a]$ ) or $b\nsim a$ (hence, $b\neq a$ ) and $b<^{\mathfrak{A}}a$ ($[b]\prec[a]$ ).
3. Transitivity. For every $a$ , $b$ and $c$ , if $a\nsim b$ and $a<^{\mathfrak{A}}b$ ($[a]\prec[b]$ ) and $b\nsim c$ and $b<^{\mathfrak{A}}c$ ($[b]\prec[c]$ ), then $a<^{\mathfrak{A}}c$ and $a\nsim c$ ($[a]\prec[c]$ ). Here, we have $a\nsim c$ , as otherwise, $a<^{\mathfrak{A}}b<^{\mathfrak{A}}c=\mathbb{S}^{n}a$ , and $b\nsim a$ , implying for some $0\le k<n$ , $\mathbb{S}^{k}a<^{\mathfrak{A}}b<^{\mathfrak{A}}\mathbb{S}^{k+1}a$ , and $b\le^{\mathfrak{A}}\mathbb{S}^{k}a$ (by L1).