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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(add2(n, x), y) -> APP2(x, y)
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(x1)
add2(x1, x2)  =  add1(x2)

Lexicographic Path Order [19].
Precedence:
[APP1, add1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
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
                ↳ QDPOrderProof
              ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


REVERSE1(add2(n, x)) -> REVERSE1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REVERSE1(x1)  =  REVERSE1(x1)
add2(x1, x2)  =  add2(x1, x2)

Lexicographic Path Order [19].
Precedence:
add2 > REVERSE1


The following usable rules [14] were oriented: none



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