However, in all examples below on how to implement MCPs, the variable z seems to be reduced to a variable with only an upper bound OR a lower bound. In that case, the code goes something like
z.lo = 3;
…
eq_1 … F(z) =g= 0;
…
model example eq_1.z ;

I don’t see a statement anywhere that tells me how to implement a model where F(z) can be both ≤0 and ≥0, depending on the bound that z reaches. Moreover, I’m told in Table 2 that double bound variables can only be matched with =N= equations, and I don’t think that’s what I want (I want F(Z) to be equal to zero in all cases where z is between its bounds, so an inequality sign seems wrong to me).
I wasn’t able to find an example for code written for an MCP, where a Variable has an upper and a lower bound, resulting in three possible stages.

Let’s take a simple MCP as an example. Let’s start with an optimization problem, so we have some intuition from the optimization world we can lean on.

min f(x) := sqr(x-1) s.t. L <= x <= U, where L and U can be finite or the expected infinity.

Taking the KKT conditions we get this MCP:
F(x) := 2(x-1) perp to L <= x <= U

I’ve attached a GAMS version of this. Try this with all the interesting combinations: L = -INF or finite and less than 1 or 1 or greater than 1, similar for for U. That’s 16 combinations. Try them all. If you understand this tiny example and each combination, I think you’ll be in good shape regarding this part of MCP. mcp.gms (343 Bytes)
To make the experiment explicit: L in {-inf, 0, 1, 2}, U in {0, 1, 2, +INF} yields 16 combinations.

Thank you for your reply.
I suppose you meant to set the initial goal to min f(x) := (x-1)**2. I ran it with your suggested domains and some additional ones, such as [-2;0] or [2;5], to see which solution the model chooses if neither x.lo, nor x.up, nor anything in between is able to solve the equation with an equality sign. That was actually a very helpful exercise for me - thank you!

So, the short answer is: NE does create these three stages. Nice!