A non-linear formulation is given for $ x \geq 0$, $ x’ \geq 0$, $z \geq 0$ and $ \beta $ binary.
$\left\{ \begin{array}{l}
x \geq \beta \cdot z \\
x’ \geq (1- \beta) \cdot z \\
\end{array} \right. $
If $ \beta = 1 $ we get
$\left\{ \begin{array}{l}
x \geq z \\
x’ \geq 0 \\
\end{array} \right. $
If $ \beta = 0 $ we get
$\left\{ \begin{array}{l}
x \geq 0 \\
x’ \geq z \\
\end{array} \right. $
Oppositive case
$\left\{ \begin{array}{l}
z \geq \beta \cdot x \\
z \geq (1- \beta) \cdot x’ \\
\end{array} \right. $
The following formulation is linear, but it is valid only for $z < 1$.
$\left\{ \begin{array}{l}
x \geq (\beta -1) + z \\
x’ \geq - \beta + z \\
\end{array} \right. $
As known, $ (0;1) \cong R$ meaning that a bijective function exists between the open interval and the set of real numbers. Also the generic interval $ (a;b) $ is isomorph to $ (0;1) $, $ (a;b) \cong (0;1) $ by means of the following bijection:
$ f(x):= (x-a)/(b-a) $
Thanks to this observation, the linear formulation of the logical constraints can be given changing the interval where variables are defined. From $ x \geq 0$, $ x’ \geq 0$, $z \geq 0$ and $ \beta $ binary, let introduce $ y := (x-a)/(b-a) $, $ y’ := (x’-a)/(b-a) $’ $ w := (z-a)/(b-a) $ where $a$ and $b$ are suggested from the context of the examinated problem.
‘> $\left\{ \begin{array}{l}
y \geq ( \beta - 1) + w \\
y’ \geq - \beta + w \\
y \in (0;1) \\
y’ \in (0;1) \\
w \in (0;1) \\
\beta binary \\
\end{array} \right. $