Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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

Q restricted rewrite system:
The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

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:

ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
PREFIX(L) → NIL
APP(cons(X, XS), YS) → ACTIVATE(XS)
ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__from(X)) → FROM(X)
PREFIX(L) → PREFIX(L)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
PREFIX(L) → NIL
APP(cons(X, XS), YS) → ACTIVATE(XS)
ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__from(X)) → FROM(X)
PREFIX(L) → PREFIX(L)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

PREFIX(L) → PREFIX(L)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ NonTerminationProof

          ↳ QDP

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

PREFIX(L) → PREFIX(L)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

PREFIX(L) → PREFIX(L)

The TRS R consists of the following rules:none

s = PREFIX(L) evaluates to t =PREFIX(L)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from PREFIX(L) to PREFIX(L).

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
APP(cons(X, XS), YS) → ACTIVATE(XS)
ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

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

ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
APP(cons(X, XS), YS) → ACTIVATE(XS)
ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

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:

ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
The graph contains the following edges 2 > 1
APP(cons(X, XS), YS) → ACTIVATE(XS)
The graph contains the following edges 1 > 1
ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
The graph contains the following edges 1 > 1
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
The graph contains the following edges 2 > 1