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, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

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(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

The set Q consists of the following terms:

app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))


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

REVERSE1(add2(n, x)) -> APP2(reverse1(x), add2(n, nil))
REVERSE1(add2(n, x)) -> REVERSE1(x)
SHUFFLE1(add2(n, x)) -> SHUFFLE1(reverse1(x))
SHUFFLE1(add2(n, x)) -> REVERSE1(x)
APP2(add2(n, x), y) -> APP2(x, y)

The TRS R consists of the following rules:

app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

The set Q consists of the following terms:

app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))

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:

REVERSE1(add2(n, x)) -> APP2(reverse1(x), add2(n, nil))
REVERSE1(add2(n, x)) -> REVERSE1(x)
SHUFFLE1(add2(n, x)) -> SHUFFLE1(reverse1(x))
SHUFFLE1(add2(n, x)) -> REVERSE1(x)
APP2(add2(n, x), y) -> APP2(x, y)

The TRS R consists of the following rules:

app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

The set Q consists of the following terms:

app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

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

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

APP2(add2(n, x), y) -> APP2(x, y)

The TRS R consists of the following rules:

app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

The set Q consists of the following terms:

app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(add2(n, x), y) -> APP2(x, y)
Used argument filtering: APP2(x1, x2)  =  x1
add2(x1, x2)  =  add1(x2)
Used ordering: Quasi Precedence: trivial


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

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

app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

The set Q consists of the following terms:

app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))

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
                ↳ QDPAfsSolverProof
              ↳ QDP

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

REVERSE1(add2(n, x)) -> REVERSE1(x)

The TRS R consists of the following rules:

app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

The set Q consists of the following terms:

app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

REVERSE1(add2(n, x)) -> REVERSE1(x)
Used argument filtering: REVERSE1(x1)  =  x1
add2(x1, x2)  =  add1(x2)
Used ordering: Quasi Precedence: trivial


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

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

app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

The set Q consists of the following terms:

app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))

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
QDP

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

SHUFFLE1(add2(n, x)) -> SHUFFLE1(reverse1(x))

The TRS R consists of the following rules:

app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
reverse1(nil) -> nil
reverse1(add2(n, x)) -> app2(reverse1(x), add2(n, nil))
shuffle1(nil) -> nil
shuffle1(add2(n, x)) -> add2(n, shuffle1(reverse1(x)))

The set Q consists of the following terms:

app2(nil, x0)
app2(add2(x0, x1), x2)
reverse1(nil)
reverse1(add2(x0, x1))
shuffle1(nil)
shuffle1(add2(x0, x1))

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