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, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

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, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

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)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ZWADR2(XS, YS)
PREFIX1(L) -> ZWADR2(L, prefix1(L))
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> APP2(Y, cons2(X, nil))
FROM1(X) -> FROM1(s1(X))
APP2(cons2(X, XS), YS) -> APP2(XS, YS)

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

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)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ZWADR2(XS, YS)
PREFIX1(L) -> ZWADR2(L, prefix1(L))
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> APP2(Y, cons2(X, nil))
FROM1(X) -> FROM1(s1(X))
APP2(cons2(X, XS), YS) -> APP2(XS, YS)

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 2 less nodes.

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

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

FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

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

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

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

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

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

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.


APP2(cons2(X, XS), YS) -> APP2(XS, YS)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( cons2(x1, x2) ) = 2x2 + 1


POL( APP2(x1, x2) ) = x1 + 3



The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ZWADR2(XS, YS)

The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

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)) -> ZWADR2(XS, YS)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ZWADR2(x1, x2) ) = 3x2 + 2


POL( cons2(x1, x2) ) = 2x1 + 2x2 + 3



The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))

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