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(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

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:

APP2(asort, z) -> APP2(sort, min)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(app2(sort, f), g), y)
APP2(dsort, z) -> APP2(sort, max)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
APP2(dsort, z) -> APP2(app2(sort, max), min)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(app2(insert, f), g), z)
APP2(app2(app2(app2(insert, f), g), nil), y) -> APP2(cons, y)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(g, x), y)
APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(app2(min, x), y)
APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(app2(max, x), y)
APP2(asort, z) -> APP2(app2(app2(sort, min), max), z)
APP2(app2(app2(app2(insert, f), g), nil), y) -> APP2(app2(cons, y), nil)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(insert, f), g)
APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(max, x)
APP2(asort, z) -> APP2(app2(sort, min), max)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(f, x), y)
APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(min, x)
APP2(dsort, z) -> APP2(app2(app2(sort, max), min), z)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(f, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y))
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y))
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(g, x)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(insert, f)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(cons, app2(app2(f, x), y))

The TRS R consists of the following rules:

app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

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:

APP2(asort, z) -> APP2(sort, min)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(app2(sort, f), g), y)
APP2(dsort, z) -> APP2(sort, max)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
APP2(dsort, z) -> APP2(app2(sort, max), min)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(app2(insert, f), g), z)
APP2(app2(app2(app2(insert, f), g), nil), y) -> APP2(cons, y)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(g, x), y)
APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(app2(min, x), y)
APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(app2(max, x), y)
APP2(asort, z) -> APP2(app2(app2(sort, min), max), z)
APP2(app2(app2(app2(insert, f), g), nil), y) -> APP2(app2(cons, y), nil)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(insert, f), g)
APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(max, x)
APP2(asort, z) -> APP2(app2(sort, min), max)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(f, x), y)
APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(min, x)
APP2(dsort, z) -> APP2(app2(app2(sort, max), min), z)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(f, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y))
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y))
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(g, x)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(insert, f)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(cons, app2(app2(f, x), y))

The TRS R consists of the following rules:

app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 14 less nodes.

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

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

APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(app2(min, x), y)

The TRS R consists of the following rules:

app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

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 strictly oriented and are deleted.


APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(app2(min, 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)  =  x1
app2(x1, x2)  =  app2(x1, x2)
min  =  min
s  =  s

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ 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(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

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
            ↳ QDPOrderProof
          ↳ QDP

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

APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(app2(max, x), y)

The TRS R consists of the following rules:

app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

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 strictly oriented and are deleted.


APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(app2(max, 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)  =  x1
app2(x1, x2)  =  app2(x1, x2)
max  =  max
s  =  s

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ 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(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

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

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

APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(g, x), y)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y))
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(app2(sort, f), g), y)
APP2(asort, z) -> APP2(app2(app2(sort, min), max), z)
APP2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> APP2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
APP2(dsort, z) -> APP2(app2(app2(sort, max), min), z)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(app2(f, x), y)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(g, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, x), y)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), y)), x)
app2(app2(app2(app2(insert, f), g), nil), y) -> app2(app2(cons, y), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, x), z)), y) -> app2(app2(cons, app2(app2(f, x), y)), app2(app2(app2(app2(insert, f), g), z), app2(app2(g, x), y)))
app2(app2(max, 0), y) -> y
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), y) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(asort, z) -> app2(app2(app2(sort, min), max), z)
app2(dsort, z) -> app2(app2(app2(sort, max), min), z)

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