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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 $a\dot{=}b\dot{=}\ldots$ ?

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    You need to restate your freflexivity statement; the pair of lines don't make much sense together. And for your measure of equality statement, as stated, this would be satisfied by the $\leq$ relation but I suspect (from your antisymmetry axiom) you meant that the $<$ relation would be OK as well. – 2017-01-30
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    Something is confusing here: You write two statements for reflecivity, the second is trivially true. And Antisymmetry is also trivially true by definition of $\dot=$. And successor is trivially true by the first version of reflexivity. And immediate predecessors cannot possibly exist. I suppose you want to replace a few $\prec$ with $\preceq$ perhaps? – 2017-01-30
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    @HagenvonEitzen yes sorry I've stated anti symmetry and defined dot equality simply because I wanted to clearly discriminate dot equality from equality – 2017-01-30
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    Antisymmetry doesn't imply a=b it only implies a dot-equals b. – 2017-01-30
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    @HagenvonEitzen you could replace it with any symbol you wanted but I wanted to be explicit which equals the successor or dot-equal referred to. – 2017-01-30
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    The first part just seems like what you want is a partial preorder with the dot-equality being the associated equivalence relation. – 2017-01-30
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    @MarkFischler I have been quite careful. I think it reflects my intention correctly. – 2017-01-30
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    @TobiasKildetoft you're talking a foreign language to me but that does sound right because this structure can be described by a partially ordered sequence of equivalence classes. The question is equivalent to whether they are totally ordered in succession towards a single class containing a single element. Actually... typing this I now realise there is an error in this :( – 2017-01-30
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    @TobiasKildetoft ok fixed it. If there is a way I think it's Zorn's lemma but it's beyond me. – 2017-01-30
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    (1).For all $n\in \mathbb Z^+$ you have $(n \succ n \lor (n\dot {=}n))$ which is equivalent to $(n\succ n\lor (n\succ n\lor n\succ n)$ which is equiv. to $n\succ n$...(2). The Antisymmetry statement is equivalent to the def'n of $a\dot {=}b$ ....(3). Since by my (1),we have $n\succ n$ for all $n$, the unique predecessor axiom is false because $ n\succ n\succ n.$ – 2017-01-30
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    @user254665 fabulous work, thank- you! Although you may notice this goes against the spirit of the unique predecessor rule! I will fix it. – 2017-01-30
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    @user254665 there you go. Fixed the unique predecessor; I think. – 2017-01-30
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    @TobiasKildetoft I have read up and yes, a partial pre-order is exactly what I have attempted to describe, so thanks for that. – 2017-01-31
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    As @MarkFischler already said, your "reflexivity" statements makes no sense, so why haven't you fixed it? And your anti-symmetry axiom is bloated; it trivially reduces to "$\forall a,b\ ( a \prec b \prec a \to a \overset{\cdot}= b )$" by basic logic. Zorn's lemma has **absolutely nothing** to do with your question. Zorn's lemma is only relevant when you have **uncountable** sets, otherwise DC (dependent choice) is all you ever need. – 2017-01-31
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    But at least you got your "transitivity" and "predecessor" axioms to make sense. The rest don't quite make sense. – 2017-01-31

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