- * Sort Parameters first according to type then to id and lastly to channel.
- *
- * Proof:
- * <ol>
- * <li>Irreflexivity: f(x, x) is false because of the irreflexivity of \c < in each branch.</li>
- * <li>Antisymmetry: given x != y, f(x, y) implies !f(y, x) because of the same
- * property of \c < in each branch and the symmetry of operator==. </li>
- * <li>Transitivity: let f(x, y) and f(y, z) => f(x, z) be true.
- * We prove by contradiction, assuming the contrary:
- * f(x, y) and f(x, z) hold => !f(x, z)
- *
- * That implies one of the following:
- * <ol>
- * <li> x == z which contradicts the assumption f(x, y) and f(y, x)
- * because of antisymmetry.
- * </li>
- * <li> f(z, x) is true. That would imply that one of the ivars (we call it i)
- * of x is greater than the same ivar in z while all "previous" ivars
- * are equal. That would imply that also in y all those "previous"
- * ivars are equal and because if x.i > z.i it is impossible
- * that there is an y that satisfies x.i < y.i < z.i at the same
- * time which contradicts the assumption.
- * </li>
- * Therefore f(x, z) is true (transitivity)
- * </ol>
- * </li>
- * </ol>