Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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.

(5) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(6) Obligation:

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.

(7) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))

(8) Obligation:

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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

• R(xs, nil, zs, cons(w, ws)) → R(xs, xs, cons(succ(zero), zs), ws)
  The graph contains the following edges 1 >= 1, 1 >= 2, 4 > 4

• R(xs, cons(y, ys), cons(z, zs), cons(w, ws)) → R(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws)))
  The graph contains the following edges 2 > 1, 2 >= 2, 3 > 3

• R(xs, cons(y, ys), nil, cons(w, ws)) → R(xs, xs, cons(succ(zero), nil), ws)
  The graph contains the following edges 1 >= 1, 1 >= 2, 4 > 4

(10) TRUE