What is the reason to stipulate $0 \neq 1$ in the definition of Ordered Integral Domains?

Question

Im using the book Number Systems and Foundations of Analysis by Elliot Mendelson.

The author asked in the book which structures we are removing when stipulating that $0 \neq 1$, the first thing which come to my mind is that this condition just remove all structures which have an unitary underlying set, but I think is probably something Im missing, I know there are other questions talking about the zero ring being or not a integral domain or ring, but here in this case the author consider the zero ring even as an integral domain.

But the zero ring is ruled out of ordered integral domains because the author stipulates that $0 \neq 1$ in the definition of the order relation Axioms

There is something special about this stipulation? We are removing more kind of structures than I have noticed? and if we dont use this stipulation it will cause trouble in other definitions?

Here I will put a copy of the Axioms used in the book to define the structures:

Definitions:

By a ring we mean a triple $(R,+,\times)$, such that $R$ is a set and $+$ and $\times$ are binary operations of $R$ satisfying:

$+$ is associative
$+$ is commutative
There is an element $0$ in $R$ such that

3.1 $(\forall x)(x+0 = x)$
3.2 $(\forall x)(\exists y)(x+y = 0)$

$\times$ is associative
$\times$ distributes over $+$

By a ring with unit element, we mean a ring $(R,+,\times)$:

There is an element $1$ such that $x=x \times 1 = 1 \times x$, for all $x$ in $R$.

By a commutative ring, we mean a ring $(R,+,\times)$ where:

$\times$ is commutative

By an integral domain we mean a commutative ring with unit element $(R,+,\times)$ satisfying the addtional axiom:

$[x \neq 0 \land y \neq 0] \Rightarrow x \times y \neq 0$ for any $x$ and $y$ in $R$

By an ordered integral domain we mean a quadruple $(R,+,\times,\lt)$, consisting of an integral domain $(R,+,\times)$ together with a binary relation $\lt$ in $R$ satisfying the following Axioms:

(O1) $\lnot(x \lt x)$

(O2) $[x \lt y \land y \lt z] \Rightarrow x \lt z$

(O3) $[x \lt y \lor x=y \lor y \lt x]$

(O4) $x \lt y \Rightarrow x+z \lt y+z$

(O5) $[x \lt y \land 0 \lt z] \Rightarrow x \times z \lt y \times z$

In addition we stipulate that $0 \neq 1$.

If $0=1$ you reduce to one element (the distributive law kills everything even before the order axioms). This is a valid (even interesting) object of study, but often ends up being an exception in the development of the theory - so best to study it separately and exclude it from the more general theory, which is then developed without exception. — Mark Bennet, Oct 18 '20 at 20:23
and also Explanation for why $1\neq 0$ is explicitly mentioned in Chapter 1 of Spivak's Calculus for properties of numbers. — Bill Dubuque, Oct 18 '20 at 20:54
I dont have studied the field definition yet, but I if i understood the other threads this convention prevent the trivial ring in fields and integral domains because it will need to be stated in every theorem, and as you stated in your answer in the other thread it is useful to establish proofs by contradiction in some cases.
I just wonder why Elliot Mendelson allowed the the trivial case in integral domains, his stipulation of $0 \neq 1$ just appears when talking about ordered integral domains. — Paulo Henrique L. Amorim, Oct 19 '20 at 12:15

What is the reason to stipulate $0 \neq 1$ in the definition of Ordered Integral Domains?

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