Termination w.r.t. Q of the following Term Rewriting System could be proven:

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 [19] 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 [1,15] 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)))

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
          ↳ UsableRulesProof

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)))

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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)))

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPSizeChangeProof

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)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: