How to formulate x != y in lpsolve?

I'm trying to formulate that the variables x,y,z must all be different and that they accept only the values 1, 2 or 3 (this is, of course, a toy example):

min: x+y+z;

1 <= x <= 3;
1 <= y <= 3;
1 <= z <= 3;

but to make this work I still need either access to boolean operators or to a != operator, which don't seem to exist in lpsolve! How can I go about this? I wanted to do this:

x != y;
x != z;
y != z;

Thanks

EDIT:

Here's my current code:

/* Objective function */
min: 1;

/* Variable bounds */
1 <= x1 <= 4;
1 <= x2 <= 4;
1 <= x3 <= 4;
1 <= x4 <= 4;


x1 + x2 + x3 + x4 = 10;

x1 < x2;
x2 < x3;
x3 < x4;

int x1;
int x2;
int x3;
int x4;

lpsolve is giving me as result:

x1 = 1
x2 = 3
x3 = 3
x4 = 3

which is wrong. Why?

In general I would agree with Michael Laffargue that it is not possible to have something like a < b for real a,b in lpSolve. But for integer expressions that is a bit different.

Maybe we can start with a more simplified problem. Let us think of two integer variables x and y and a constant M such that 1 <= x <= M and 1<=y<=M. If x and y may not be equal then x>y or y>x. But as both are integers only one of the following inequalities hold

x+1 <= y
y+1 <= x

We can enforce that only one of the inequalities above holds by introducing a binary variable r such that for x,y, r the following inequalities hold both :

(i)   x+1 <= y+ Mr
(ii)  y+1 <= x+M-Mr

because if r=0 then (i) x+1 <=y and (ii) is trivial, but if r=1 then (ii) y+1 <= x and (i) is trivial.

When we now apply the solution from above to the problem of the OP, then we can build a linear program with inequalities (i) and (ii) for all pairs of variables from the OP's problem and M=4:

/* Objective function */
min: 1;

/* Variable bounds */
1 <= x1 <= 4;
1 <= x2 <= 4;
1 <= x3 <= 4;
1 <= x4 <= 4;
r_12 <= 1;
r_13 <= 1;
r_14 <= 1;
r_23 <= 1;
r_24 <= 1;
r_34 <= 1;

/* This is done automatically because all x1,..,x4 are different
   x1 + x2 + x3 + x4 = 10;
*/

/*  Apply (i) and (ii) to all pairs of x1,..x4
(i)  x+1 <= y + Mr
(ii) y+1 <= x + M-Mr
*/

x1 + 1 <= x2 + 4 r_12;
x2 + 1 <= x1 + 4 - 4 r_12;

x1 + 1 <= x3 + 4 r_13;
x3 + 1 <= x1 + 4 - 4 r_13;

x1 + 1 <= x4 + 4 r_14;
x4 + 1 <= x4 + 4 - 4 r_14;

x2 + 1 <= x3 +  4 r_23;
x3 + 1 <= x2 + 4 - 4 r_23;

x2 + 1 <= x4 + 4 r_24;
x4 + 1 <= x2 + 4 - 4 r_24;

x3 + 1 <= x4 + 4 r_34;
x4 + 1 <= x3 + 4 - 4 r_34;

/*
x1 < x2;
x2 < x3;
x3 < x4;
*/

int r_12;
int r_13;
int r_14;
int r_23;
int r_24;
int r_34;
int x1;
int x2;
int x3;
int x4;

Solving the MILP above renders a solution with all x1,..,x4 different: