Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
r(xs, ys, zs, nil) → xs
r(xs, nil, zs, cons(w, ws)) → r(xs, xs, cons(succ(zero), zs), ws)
r(xs, cons(y, ys), nil, cons(w, ws)) → r(xs, xs, cons(succ(zero), nil), ws)
r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
r(xs, ys, zs, nil) → xs
r(xs, nil, zs, cons(w, ws)) → r(xs, xs, cons(succ(zero), zs), ws)
r(xs, cons(y, ys), nil, cons(w, ws)) → r(xs, xs, cons(succ(zero), nil), ws)
r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
R(xs, nil, zs, cons(w, ws)) → R(xs, xs, cons(succ(zero), zs), ws)
R(xs, cons(y, ys), nil, cons(w, ws)) → R(xs, xs, cons(succ(zero), nil), ws)
R(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → R(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
The TRS R consists of the following rules:
r(xs, ys, zs, nil) → xs
r(xs, nil, zs, cons(w, ws)) → r(xs, xs, cons(succ(zero), zs), ws)
r(xs, cons(y, ys), nil, cons(w, ws)) → r(xs, xs, cons(succ(zero), nil), ws)
r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
R(xs, nil, zs, cons(w, ws)) → R(xs, xs, cons(succ(zero), zs), ws)
R(xs, cons(y, ys), nil, cons(w, ws)) → R(xs, xs, cons(succ(zero), nil), ws)
R(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → R(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
The TRS R consists of the following rules:
r(xs, ys, zs, nil) → xs
r(xs, nil, zs, cons(w, ws)) → r(xs, xs, cons(succ(zero), zs), ws)
r(xs, cons(y, ys), nil, cons(w, ws)) → r(xs, xs, cons(succ(zero), nil), ws)
r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.