xor as linear inequalities

I came across a question about using an xor condition in a MIP model. Here is a summary of the answer.

Expressing $z=x\cdot y$ (or $z=x\text{ and }y$ ) where $x,y,z \in \{0,1\}$ as a set of linear constraints is well-known:

and:

$$\begin{align} &z \le x \\ &z \le y \\ &z \ge x+y-1 \end{align}$$

$\>\>\>\Longleftrightarrow\>\>\>$

$z$ $x$ $y$ $0$ $0$ $0$ $0$ $0$ $1$ $0$ $1$ $0$ $1$ $1$ $1$

The relation $z=x\text{ xor }y$ is slightly more complicated. The xor (exclusive-or) condition can also be written as $z = x<>y$, i.e. $z$ is 1 if $x$ and $y$ are different (and $z$ is zero if they are the same). This we can write as the following set of linear inequalities:

xor:

$$\begin{align} & z \le x+y\\ & z \ge x-y \\ & z \ge y-x \\ & z \le 2 -x - y \end{align}$$

$\>\>\>\Longleftrightarrow\>\>\>$

$z$ $x$ $y$ $0$ $0$ $0$ $1$ $0$ $1$ $1$ $1$ $0$ $0$ $1$ $1$

To be complete $z=x\text{ or }y$ can be written as:

or:

$$\begin{align} & z \le x+y\\ & z \ge x \\ & z \ge y \end{align}$$

$\>\>\>\Longleftrightarrow\>\>\>$

$z$ $x$ $y$ $0$ $0$ $0$ $1$ $0$ $1$ $1$ $1$ $0$ $1$ $1$ $1$

In all of this we assumed $x,y$ and $z$ are binary variables.