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, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)


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)
ZWADR(cons(X, XS), cons(Y, YS)) → ZWADR(XS, YS)
PREFIX(L) → ZWADR(L, prefix(L))
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, nil))
FROM(X) → FROM(s(X))
APP(cons(X, XS), YS) → APP(XS, YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

PREFIX(L) → PREFIX(L)
ZWADR(cons(X, XS), cons(Y, YS)) → ZWADR(XS, YS)
PREFIX(L) → ZWADR(L, prefix(L))
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, nil))
FROM(X) → FROM(s(X))
APP(cons(X, XS), YS) → APP(XS, YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

ZWADR(cons(X, XS), cons(Y, YS)) → ZWADR(XS, YS)
PREFIX(L) → PREFIX(L)
PREFIX(L) → ZWADR(L, prefix(L))
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, nil))
APP(cons(X, XS), YS) → APP(XS, YS)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

APP(cons(X, XS), YS) → APP(XS, YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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.


APP(cons(X, XS), YS) → APP(XS, YS)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
cons(x1, x2)  =  cons(x2)

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

Status:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

ZWADR(cons(X, XS), cons(Y, YS)) → ZWADR(XS, YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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)) → ZWADR(XS, YS)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ZWADR(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

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

Status:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ 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, app(XS, YS))
from(X) → cons(X, from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, nil)), zWadr(XS, YS))
prefix(L) → cons(nil, zWadr(L, prefix(L)))

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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