Termination w.r.t. Q proof of TRS/TRCSREx3_3_25_Bor03

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:

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

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

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

      ↳ EdgeDeletionProof

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

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

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 deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

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 [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

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

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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__app(X1, X2)) → APP(X1, X2)

Used ordering: Combined order from the following AFS and order.
ZWADR(x1, x2) = ZWADR(x1, x2)
cons(x1, x2) = cons(x1, x2)
ACTIVATE(x1) = x1
n__zWadr(x1, x2) = n__zWadr(x1, x2)
n__app(x1, x2) = n__app(x1, x2)
APP(x1, x2) = APP(x1, x2)
n__nil = n__nil

Recursive path order with status [2].
Quasi-Precedence:

cons₂ > [ZWADR₂, n_{_app₂}, APP_2] > n_{_nil}
n_{_zWadr₂} > [ZWADR₂, n_{_app₂}, APP_2] > n_{_nil}

Status:

n_{_zWadr₂}: multiset
n_{_nil}: multiset
n_{_app₂}: multiset
APP₂: multiset
ZWADR₂: multiset
cons₂: [2,1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

ACTIVATE(n__app(X1, X2)) → APP(X1, X2)

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 [13,14,18] contains 0 SCCs with 1 less node.