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:

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 oriented strictly and are deleted.

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)
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 be oriented weakly.

APP₂(cons₂(X, XS), YS) -> ACTIVATE₁(XS)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( n__nil ) = 1

POL( n__zWadr₂(x₁, x₂) ) = 3x₁ + 3x₂ + 2

POL( ACTIVATE₁(x₁) ) = max{0, 2x₁ - 3}

POL( n__app₂(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( APP₂(x₁, x₂) ) = max{0, 2x₁ - 2}

POL( ZWADR₂(x₁, x₂) ) = max{0, x₁ + 2x₂ - 2}

POL( cons₂(x₁, x₂) ) = 3x₁ + 3x₂ + 1

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:

APP₂(cons₂(X, XS), YS) -> ACTIVATE₁(XS)

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 1 less node.