Suppose some ordering $\succ$ of the positive integers $\{n\in\mathbb{N_{>0}}\}$, satisfies:
Transitivity:
$a\succ b \land b\succ c \implies a\succ c$,
$\succ$ generates a measure of equality $\dot{=}$ not necessarily corresponding with $=\quad$:
$a\dot{=}b\iff a\succ b \land b\succ a$
Reflexivity:
$n \succ n \lor n \dot{=} n \quad \forall n \in \mathbb{N_{>0}}$
$n \succ n \lor n=n \quad \forall n \in \mathbb{N_{>0}}$
Antisymmetry:
$(a\succ b\lor a\dot{=}b) \land (b\succ a\lor b\dot{=}a) \implies a\dot{=}b$
Every element has a successor:
$\forall n\in\mathbb{N_{>0}} \quad\exists\quad m: m\succ n$
Every element has a unique immediate predecessor:
$\forall n\in\mathbb{N_{>0}} \quad\exists m:(n\succ m \quad \land \quad \nexists p:(n\succ p\succ m\land p\dot{\neq}n\land p\dot{\neq}m ))$
Every totally ordered subset of $\mathbb{N_{>0}}$ contains at least one minimal element:
$Q=\{q:(n\succ q \quad \forall n\dot{\neq} q)\}$. $Q\neq\emptyset$
^ I struggled to write this; help me out if you can.
Can you show that there is only one set of elements satisfying $a\dot{=}b\dot{=}\ldots$ ?