Let $\mathbb{K}$ be a field. A subset $\mathbb{K}^{+}$ is an ordered field if: $x$ is either $x∈\mathbb{K}^{+}$, $x=0$, or $−x∈\mathbb{K}^{+}$. Secondly, if $x,y∈\mathbb{K}^{+}$, then $x+y,x⋅y∈\mathbb{K}^{+}$.
Prove $x=y ⇔ x≤y ∧ x≥y$ with $x,y \in \mathbb{K}$, if $ \mathbb{K}$ is an ordered field.
For "$\Rightarrow$"
Since $x=y$, it is $x\geq x$ and $x\leq x$ which means by definition that:
$x\geq x$, if $x-x\in \mathbb{K}^+\cup \{0\}$ and
$x\geq x$, if $x-x\in \mathbb{K}^+\cup \{0\}$
which is true.
Is that correct and how do I do the other way "$\Leftarrow$"
For the other direction you have $y-x \in \Bbb K^+ \cup {0}$ and $x-y=-(y-x) \in \Bbb K^+ \cup {0}$. You should be able to show that it must be that $y-x=0$