re-writing a $\min(X,Y)$ function linearly for LP problem I am trying to formulate an LP problem.  In the problem I have a $\min(X,Y)$ that I would like to formulate linearly as a set of constraints.  For example, replacing $\min(X,Y)$ with some variable $Z$, and having a set of constraints on Z.
I believe that there are a minimum of two constraints:
subto: $Z \le X$
subto: $Z \le Y$
That will make it take a value that is less than or equal to $\min(X,Y)$.  But, I want it to take the minimum value of $X$ or $Y$.  I am missing one constraint, that seems to have the following logic: "$Z \ge X$ or $Z \ge Y$" ... so that it isn't just less than the minimum, it IS the minimum.  I know I'm missing something basic.

In addition to fabee's response, I also have found this representation to work well which uses  either-or constraint representation.  Note that M must be large, see 15.1.2.1 in this document.
param a := 0.8;
param b := 0.4;
param M := 100;
var z;
var y binary;

minimize goal: z;

subto min_za:
  z <= a;

subto min_zb:
  z <= b;

subto min_c1:
  -z <= -a + M*y;

subto min_c2:
  -z <= -b + M*(1-y);


You could use $\min(x,y) = \frac{1}{2}(x + y - |x - y|)$ where $|x - y|$ can be replaced by the variables $z_1 + z_2$ with constraints $z_i \ge 0$ for $i=1,2$ and $z_1 - z_2 = x - y$. $z_1$ and $z_2$ are, therefore, the positive or the negative part of $|x-y|$.
Edit: 
For the reformulation to work, you must ensure that either $z_1=0$ or $z_2=0$ at the optimum, because we want 
$$z_1 = \begin{cases}
x-y & \mbox{ if }x-y\ge0\\
0 & \mbox{ otherwise}
\end{cases}$$
and 
$$z_2 = \begin{cases}
y-x & \mbox{ if }x-y\le0\\
0 & \mbox{ otherwise}
\end{cases}.$$
You can check that the constraints will be active if the objective function can always be increase/decreased by making one of the $z_i$ smaller. That is the reason why your maximization worked, because the objective function could be increased by making one of the $z_i$ smaller. 
You could fix your minimization example by requiring that $0\le z_i \le |x-y|$. In that case, the objective function will be smallest if $z_i$ are largest. However, since they still need to be $3$ apart, one of the $z_i$ will be three and the other one will be zero. 
