Do the axioms of ordered field imply that $a\cdot 0=0$ and $0<1$? We introduce some definitions:

A structure $\mathfrak{A}=\langle A,<,+,\cdot,0,1 \rangle$ where $<$ is a linear ordering, $+$ and $\cdot$ are binary operations, and $0,1$ are constants such that all properties 1-12 are satisfied is called an ordered field.
For all $a,b,c\in A$:


*

*$a+b=b+a$.

*$(a+b)+c=a+(b+c)$.

*$a+0=a$.

*There exists $a'\in A$ such that $a+a'=0$. We denote $a'=-a$, the opposite of $a$.

*$a<b\implies a+c<b+c$.

*$a\cdot (b+c)=a\cdot b + a\cdot c$.

*$a\cdot b = b\cdot a$.

*$(a\cdot b)\cdot c = a\cdot (b\cdot c)$

*$a\cdot 1=a$

*For $a\neq 0$, there exists $a'\in A$ such that $a\cdot a'=1$. We denote $a'=a^{-1}$, the reciprocal of $a$.

*$a<b$ and $0<c$ $\implies a\cdot c < b\cdot c$.

*$0\neq 1$

In the familiar structure $\mathfrak{R}=\langle \Bbb R,<,+,\cdot,0,1 \rangle$ where $\Bbb R$ is the set of real numbers, it is trivial that $a\cdot 0=0$ and $0<1$.
I would like to prove that $a\cdot 0=0$ and $0<1$ hold for any ordered field. I have tried to no avail. Naturally, a question arises in my mind.

Do the axioms of ordered field 1-12 imply that $a\cdot 0=0$ and $0<1$?

Please shed me some light. Thank you for your help!

$a \cdot 0=0$ is a consequence of the ring axioms: $a \cdot 0= a \cdot (0+0)= a \cdot 0+ a\cdot 0$.
Assume $0 > 1$. Then $-1 = 0 + (-1) > 1 + (-1) =0$. Therefore, for any $a < b$, $(-1)a < (-1)b$, hence $0 = 1 \cdot a + (-1)a = a + (-1)a < b + (-1)a < b + (-1)b= 1 \cdot b + (-1)b = 0$, a contradiction.
