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:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, n__app2(activate1(XS), YS))
from1(X) -> cons2(X, n__from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, n__nil)), n__zWadr2(activate1(XS), activate1(YS)))
prefix1(L) -> cons2(nil, n__zWadr2(L, prefix1(L)))
app2(X1, X2) -> n__app2(X1, X2)
from1(X) -> n__from1(X)
nil -> n__nil
zWadr2(X1, X2) -> n__zWadr2(X1, X2)
activate1(n__app2(X1, X2)) -> app2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__zWadr2(X1, X2)) -> zWadr2(X1, X2)
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, n__app2(activate1(XS), YS))
from1(X) -> cons2(X, n__from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, n__nil)), n__zWadr2(activate1(XS), activate1(YS)))
prefix1(L) -> cons2(nil, n__zWadr2(L, prefix1(L)))
app2(X1, X2) -> n__app2(X1, X2)
from1(X) -> n__from1(X)
nil -> n__nil
zWadr2(X1, X2) -> n__zWadr2(X1, X2)
activate1(n__app2(X1, X2)) -> app2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__zWadr2(X1, X2)) -> zWadr2(X1, X2)
activate1(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:

PREFIX1(L) -> PREFIX1(L)
ACTIVATE1(n__zWadr2(X1, X2)) -> ZWADR2(X1, X2)
ACTIVATE1(n__app2(X1, X2)) -> APP2(X1, X2)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
PREFIX1(L) -> NIL
APP2(cons2(X, XS), YS) -> ACTIVATE1(XS)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__nil) -> NIL
ACTIVATE1(n__from1(X)) -> FROM1(X)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> APP2(Y, cons2(X, n__nil))

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, n__app2(activate1(XS), YS))
from1(X) -> cons2(X, n__from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, n__nil)), n__zWadr2(activate1(XS), activate1(YS)))
prefix1(L) -> cons2(nil, n__zWadr2(L, prefix1(L)))
app2(X1, X2) -> n__app2(X1, X2)
from1(X) -> n__from1(X)
nil -> n__nil
zWadr2(X1, X2) -> n__zWadr2(X1, X2)
activate1(n__app2(X1, X2)) -> app2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__zWadr2(X1, X2)) -> zWadr2(X1, X2)
activate1(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:

PREFIX1(L) -> PREFIX1(L)
ACTIVATE1(n__zWadr2(X1, X2)) -> ZWADR2(X1, X2)
ACTIVATE1(n__app2(X1, X2)) -> APP2(X1, X2)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
PREFIX1(L) -> NIL
APP2(cons2(X, XS), YS) -> ACTIVATE1(XS)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__nil) -> NIL
ACTIVATE1(n__from1(X)) -> FROM1(X)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> APP2(Y, cons2(X, n__nil))

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, n__app2(activate1(XS), YS))
from1(X) -> cons2(X, n__from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, n__nil)), n__zWadr2(activate1(XS), activate1(YS)))
prefix1(L) -> cons2(nil, n__zWadr2(L, prefix1(L)))
app2(X1, X2) -> n__app2(X1, X2)
from1(X) -> n__from1(X)
nil -> n__nil
zWadr2(X1, X2) -> n__zWadr2(X1, X2)
activate1(n__app2(X1, X2)) -> app2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__zWadr2(X1, X2)) -> zWadr2(X1, X2)
activate1(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
      ↳ DependencyGraphProof
        ↳ AND
QDP
          ↳ QDP

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

PREFIX1(L) -> PREFIX1(L)

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, n__app2(activate1(XS), YS))
from1(X) -> cons2(X, n__from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, n__nil)), n__zWadr2(activate1(XS), activate1(YS)))
prefix1(L) -> cons2(nil, n__zWadr2(L, prefix1(L)))
app2(X1, X2) -> n__app2(X1, X2)
from1(X) -> n__from1(X)
nil -> n__nil
zWadr2(X1, X2) -> n__zWadr2(X1, X2)
activate1(n__app2(X1, X2)) -> app2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__zWadr2(X1, X2)) -> zWadr2(X1, X2)
activate1(X) -> X

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPOrderProof

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

ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
ACTIVATE1(n__zWadr2(X1, X2)) -> ZWADR2(X1, X2)
ACTIVATE1(n__app2(X1, X2)) -> APP2(X1, X2)
APP2(cons2(X, XS), YS) -> ACTIVATE1(XS)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> APP2(Y, cons2(X, n__nil))

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, n__app2(activate1(XS), YS))
from1(X) -> cons2(X, n__from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, n__nil)), n__zWadr2(activate1(XS), activate1(YS)))
prefix1(L) -> cons2(nil, n__zWadr2(L, prefix1(L)))
app2(X1, X2) -> n__app2(X1, X2)
from1(X) -> n__from1(X)
nil -> n__nil
zWadr2(X1, X2) -> n__zWadr2(X1, X2)
activate1(n__app2(X1, X2)) -> app2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__zWadr2(X1, X2)) -> zWadr2(X1, X2)
activate1(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 strictly oriented and are deleted.


ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
APP2(cons2(X, XS), YS) -> ACTIVATE1(XS)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> APP2(Y, cons2(X, n__nil))
The remaining pairs can at least by weakly be oriented.

ACTIVATE1(n__zWadr2(X1, X2)) -> ZWADR2(X1, X2)
ACTIVATE1(n__app2(X1, X2)) -> APP2(X1, X2)
Used ordering: Combined order from the following AFS and order.
ZWADR2(x1, x2)  =  ZWADR2(x1, x2)
cons2(x1, x2)  =  cons2(x1, x2)
ACTIVATE1(x1)  =  x1
n__zWadr2(x1, x2)  =  n__zWadr2(x1, x2)
n__app2(x1, x2)  =  x1
APP2(x1, x2)  =  x1
n__nil  =  n__nil

Lexicographic Path Order [19].
Precedence:
[ZWADR2, cons2, nzWadr2] > nnil


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ DependencyGraphProof

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

ACTIVATE1(n__app2(X1, X2)) -> APP2(X1, X2)
ACTIVATE1(n__zWadr2(X1, X2)) -> ZWADR2(X1, X2)

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, n__app2(activate1(XS), YS))
from1(X) -> cons2(X, n__from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, n__nil)), n__zWadr2(activate1(XS), activate1(YS)))
prefix1(L) -> cons2(nil, n__zWadr2(L, prefix1(L)))
app2(X1, X2) -> n__app2(X1, X2)
from1(X) -> n__from1(X)
nil -> n__nil
zWadr2(X1, X2) -> n__zWadr2(X1, X2)
activate1(n__app2(X1, X2)) -> app2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__zWadr2(X1, X2)) -> zWadr2(X1, X2)
activate1(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 2 less nodes.