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(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The set Q consists of the following terms:

app2(rev, nil)
app2(rev, app2(app2(cons, x0), x1))
app2(app2(rev1, 0), nil)
app2(app2(rev1, app2(s, x0)), nil)
app2(app2(rev1, x0), app2(app2(cons, x1), x2))
app2(app2(rev2, x0), nil)
app2(app2(rev2, x0), app2(app2(cons, x1), x2))


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

APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(app2(rev1, y), l)
APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(rev1, y)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev1, x), l)
APP2(rev, app2(app2(cons, x), l)) -> APP2(rev2, x)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev2, x), l)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(rev2, y), l)
APP2(rev, app2(app2(cons, x), l)) -> APP2(rev1, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev2, y)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(cons, x), app2(app2(rev2, y), l))
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(cons, x)
APP2(rev, app2(app2(cons, x), l)) -> APP2(cons, app2(app2(rev1, x), l))
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The set Q consists of the following terms:

app2(rev, nil)
app2(rev, app2(app2(cons, x0), x1))
app2(app2(rev1, 0), nil)
app2(app2(rev1, app2(s, x0)), nil)
app2(app2(rev1, x0), app2(app2(cons, x1), x2))
app2(app2(rev2, x0), nil)
app2(app2(rev2, x0), app2(app2(cons, x1), x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(app2(rev1, y), l)
APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(rev1, y)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev1, x), l)
APP2(rev, app2(app2(cons, x), l)) -> APP2(rev2, x)
APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev2, x), l)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(rev2, y), l)
APP2(rev, app2(app2(cons, x), l)) -> APP2(rev1, x)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev2, y)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(cons, x), app2(app2(rev2, y), l))
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(cons, x)
APP2(rev, app2(app2(cons, x), l)) -> APP2(cons, app2(app2(rev1, x), l))
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The set Q consists of the following terms:

app2(rev, nil)
app2(rev, app2(app2(cons, x0), x1))
app2(app2(rev1, 0), nil)
app2(app2(rev1, app2(s, x0)), nil)
app2(app2(rev1, x0), app2(app2(cons, x1), x2))
app2(app2(rev2, x0), nil)
app2(app2(rev2, x0), app2(app2(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 9 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(app2(rev1, y), l)

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The set Q consists of the following terms:

app2(rev, nil)
app2(rev, app2(app2(cons, x0), x1))
app2(app2(rev1, 0), nil)
app2(app2(rev1, app2(s, x0)), nil)
app2(app2(rev1, x0), app2(app2(cons, x1), x2))
app2(app2(rev2, x0), nil)
app2(app2(rev2, x0), app2(app2(cons, x1), x2))

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.


APP2(app2(rev1, x), app2(app2(cons, y), l)) -> APP2(app2(rev1, y), l)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app1(x2)
rev1  =  rev1
cons  =  cons

Lexicographic Path Order [19].
Precedence:
APP1 > app1
cons > app1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The set Q consists of the following terms:

app2(rev, nil)
app2(rev, app2(app2(cons, x0), x1))
app2(app2(rev1, 0), nil)
app2(app2(rev1, app2(s, x0)), nil)
app2(app2(rev1, x0), app2(app2(cons, x1), x2))
app2(app2(rev2, x0), nil)
app2(app2(rev2, x0), app2(app2(cons, x1), x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

APP2(rev, app2(app2(cons, x), l)) -> APP2(app2(rev2, x), l)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(app2(rev2, y), l)
APP2(app2(rev2, x), app2(app2(cons, y), l)) -> APP2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The TRS R consists of the following rules:

app2(rev, nil) -> nil
app2(rev, app2(app2(cons, x), l)) -> app2(app2(cons, app2(app2(rev1, x), l)), app2(app2(rev2, x), l))
app2(app2(rev1, 0), nil) -> 0
app2(app2(rev1, app2(s, x)), nil) -> app2(s, x)
app2(app2(rev1, x), app2(app2(cons, y), l)) -> app2(app2(rev1, y), l)
app2(app2(rev2, x), nil) -> nil
app2(app2(rev2, x), app2(app2(cons, y), l)) -> app2(rev, app2(app2(cons, x), app2(app2(rev2, y), l)))

The set Q consists of the following terms:

app2(rev, nil)
app2(rev, app2(app2(cons, x0), x1))
app2(app2(rev1, 0), nil)
app2(app2(rev1, app2(s, x0)), nil)
app2(app2(rev1, x0), app2(app2(cons, x1), x2))
app2(app2(rev2, x0), nil)
app2(app2(rev2, x0), app2(app2(cons, x1), x2))

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