I'm not exactly sure what my question is, but I've observed something about orders and would like to see if there is more of a story to this.
Let's assume $\;\lt\;$ is an order on some unspecified domain, and let's define $$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} \tag{0} a \le b \;\equiv\; a \lt b \lor a = b $$ Now, as I recently discovered (when writing answer https://math.stackexchange.com/a/2251407/11994 where I used $\Ref{3}$), if the domain were the reals or the rationals then the following would hold: \begin{align} \tag{1} \langle \forall z :: z \le a \then z \le b \rangle \;\equiv\; a \le b \\ \tag{2} \langle \forall z :: z \lt a \then z \lt b \rangle \;\equiv\; a \le b \\ \tag{3} \langle \forall z :: z \lt a \then z \le b \rangle \;\equiv\; a \le b \\ \tag{4} \langle \forall z :: z \le a \then z \lt b \rangle \;\equiv\; a \lt b \\ \end{align} Now visually it looks like $\Ref{4}$ is the odd one out, but let's see if that intuition is correct: I was wondering what kind of order we need to prove each of the above.
First let's look at the $\;\when\;$ direction: there, for all 4 equivalences, we only need the fact that $\;\lt\;$ is transitive. For example, for $\Ref{3}$ we have that $\;z \lt a\;$ and $\;a \le b\;$ imply $\;z \lt b\;$ by transitivity.
For the $\;\then\;$ direction of $\Ref{1}$ and $\Ref{4}$ we see that we can simply choose $\;z := a\;$, and use the fact that $\;\le\;$ is reflexive (by definition $\Ref{0}$), in other words $\;a \le a\;$.
For $\Ref{2}$ we can choose $\;z := b\;$, and assuming $\;\lt\;$ is irreflexive (so $\;\lnot(b \lt b)\;$) we get $\;\lnot(b \lt a)\;$, and further assuming $\;\lt\;$ is trichotomous (so $\;a \lt b \lor a = b \lor b \lt a\;$) we get $\;a \le b\;$.
Finally the $\;\then\;$ direction of $\Ref{3}$ is the most complex one. The best approach I found was to go existential: $$\calc \langle \forall z :: z \lt a \then z \le b \rangle \op\then\hint{see below, where $\;\lt\;$ is transitive and irreflexive} \langle \forall z :: z \lt a \then \lnot (b \lt z) \rangle \op\equiv\hint{logic: DeMorgan} \lnot \langle \exists z :: z \lt a \land b \lt z \rangle \op\then\hint{$\;\lt\;$ is dense, so $\;b \lt a\;$ implies there is a $\;z\;$ between them} \lnot (b \lt a) \op\then\hint{$\;\lt\;$ is trichotomous} a \le b \endcalc$$ where the first step above is proved as follows: $$\calc z \le b \;\then\; \lnot(b \lt z) \op\equiv\hint{logic} \lnot(z \le b \land b \lt z) \op\when\hint{$\;\lt\;$ is transitive} \lnot(z \lt z) \op\equiv\hint{$\;\lt\;$ is irreflexive} \true \endcalc$$ So here we require that $\;\lt\;$ is transitive, irreflexive, trichotomous, and dense.
To summarize the requirements on $\;\lt\;$ that seem necessary to prove the statements above: $$\begin{array}{cl} \Ref{1}&\text{transitive}\\ \Ref{2}&\text{transitive, irreflexive, trichotomous}\\ \Ref{3}&\text{transitive, irreflexive, trichotomous, dense}\\ \Ref{4}&\text{transitive}\\ \end{array}$$
So by this table $\Ref{3}$ is the odd one out, since it requires the most properties of the order $\;\lt\;$: specifically it requires density, which makes it true for the reals and the rationals, but false for e.g. the integers.
So I think my questions are: Are all of the above indeed necessary, or can some be weakened? And is there something more to this pattern? Why does density only play a role in one of these equivalences? Is there perhaps a more symmetrical way of organizing this information?
Assume transitivity.
As you've already made the case, lifting density will give you a straightforward counter-example for $(3) (\implies)$.
To lift trichotomy, one must lift one or more of anti-reflexivity, anti-symmetry, or completeness.
Lifting completeness gives a straighforward lattice counter-example for $(2) (\implies)$, so completeness is necessary for $(2) (\implies)$.
Lifting anti-reflexivity, take a domain with some $a, b$ such that $b < b < a \land \lnot (a < a) \land \lnot (a = b) \land (\forall z: z \ge a \lor z \le b)$. Thus, $z \le b \implies z < b$, so $\forall z: z < a \implies z \le b \implies z < b$, yet $a > b \land \lnot (a = b)$, i.e. $\lnot (a \le b)$ -- so anti-reflexivity is necessary for $(2) (\implies)$ as well.
Anti-reflexivity together with transitivity implies anti-symmetry, so anti-symmetry is also necessary for $(2) (\implies)$.
With all of anti-reflexivity, anti-symmetry ans completeness necessary for $(2) (\implies)$, trichotomy is also necessary. Because $LHS (2) \implies LHS (3)$, any premise necessary for $(2) (\implies)$ is also necessary for $(3) (\implies)$, so trichotomy is also necessary for $(3) (\implies)$.
In conclusion: assuming transitivity, all the other premises are necessary as listed.
(Not sure about the rest of your questions, but you may want to check if 42 fits...)