Termination w.r.t. Q proof of TRS/ThiemannsizeChange.trs

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

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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)))

The set Q consists of the following terms:

r(x0, x1, x2, nil)
r(x0, nil, x1, cons(x2, x3))
r(x0, cons(x1, x2), nil, cons(x3, x4))
r(x0, cons(x1, x2), cons(x3, x4), cons(x5, x6))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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)))
R(xs, nil, zs, cons(w, ws)) → R(xs, xs, cons(succ(zero), zs), 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)))

The set Q consists of the following terms:

r(x0, x1, x2, nil)
r(x0, nil, x1, cons(x2, x3))
r(x0, cons(x1, x2), nil, cons(x3, x4))
r(x0, cons(x1, x2), cons(x3, x4), cons(x5, x6))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

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)))
R(xs, nil, zs, cons(w, ws)) → R(xs, xs, cons(succ(zero), zs), 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)))

The set Q consists of the following terms:

r(x0, x1, x2, nil)
r(x0, nil, x1, cons(x2, x3))
r(x0, cons(x1, x2), nil, cons(x3, x4))
r(x0, cons(x1, x2), cons(x3, x4), cons(x5, x6))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

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)))
R(xs, nil, zs, cons(w, ws)) → R(xs, xs, cons(succ(zero), zs), 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)))

The set Q consists of the following terms:

r(x0, x1, x2, nil)
r(x0, nil, x1, cons(x2, x3))
r(x0, cons(x1, x2), nil, cons(x3, x4))
r(x0, cons(x1, x2), cons(x3, x4), cons(x5, x6))

We have to consider all minimal (P,Q,R)-chains.