### Linearity and Nonlinear Residues

The clp(Q,R) system is restricted to deal with linear constraints because
the decision algorithms for general nonlinear constraints are prohibitively
expensive to run. If you need this functionality badly, you should look into
symbolic algebra packages.
Although the clp(Q,R) system cannot solve nonlinear constraints, it will
collect them faithfully in the hope that through the addition of further (linear)
constraints they might get simple enough to solve eventually.
If an answer contains nonlinear constraints, you have to be aware of the
fact that success is qualified modulo the existence of a solution to the
system of residual (nonlinear) constraints:

clp(r) ?- `{sin(X) = cos(X)}.`
clpr:{sin(X)-cos(X)=0.0}

There are indeed infinitely many solutions to this constraint
(`X = 0.785398 + n*Pi`

), but clp(Q,R) has no direct means to find
and represent them.
The systems goes through some lengths to recognize linear expressions
as such. The method is based on a normal form for multivariate
polynomials. In addition, some simple isolation axioms, that can be
used in equality constraints, have been added.
The current major limitation of the method is that full polynomial
division has not been implemented.
Examples:

This is an example where the isolation axioms are sufficient to determine
the value of `X`.

clp(r) ?- `{sin(cos(X)) = 1/2}.`
X = 1.0197267436954502

If we change the equation into an inequation, clp(Q,R) gives up:

clp(r) ?- `{sin(cos(X)) < 1/2}.`
clpr:{sin(cos(X))-0.5<0.0}

The following is easy again:

clp(r) ?- `{sin(X+2+2)/sin(4+X) = Y}.`
Y = 1.0

And so is this:

clp(r) ?- `{(X+Y)*(Y+X)/X = Y*Y/X+99}.`
{Y=49.5-0.5*X}

An ancient symbol manipulation benchmark consists in rising the expression
`X+Y+Z+1`

to the 15th power:

clp(q) ?- `{exp(X+Y+Z+1,15)=0}.`
clpq:{Z^15+Z^14*15+Z^13*105+Z^12*455+Z^11*1365+Z^10*3003+...
... polynomial continues for a few pages ...
=0}

Computing its roots is another story.