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(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

app2(app2(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

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(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(f, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(f, x), h)
APP2(descending_sort, l) -> APP2(sort, max)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(app2(sort, f), g), t)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(cons, app2(app2(f, x), h))
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
APP2(descending_sort, l) -> APP2(app2(sort, max), min)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(app2(insert, f), g), t)
APP2(app2(app2(app2(insert, f), g), nil), x) -> APP2(cons, x)
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(app2(app2(app2(insert, f), g), nil), x) -> APP2(app2(cons, x), nil)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(insert, f), g)
APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(max, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(g, x), h)
APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(min, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(g, x)
APP2(ascending_sort, l) -> APP2(app2(app2(sort, min), max), l)
APP2(ascending_sort, l) -> APP2(sort, min)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h))
APP2(ascending_sort, l) -> APP2(app2(sort, min), max)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t))
APP2(descending_sort, l) -> APP2(app2(app2(sort, max), min), l)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(insert, f)

The TRS R consists of the following rules:

app2(app2(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

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(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(f, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(f, x), h)
APP2(descending_sort, l) -> APP2(sort, max)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(app2(sort, f), g), t)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(cons, app2(app2(f, x), h))
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
APP2(descending_sort, l) -> APP2(app2(sort, max), min)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(app2(insert, f), g), t)
APP2(app2(app2(app2(insert, f), g), nil), x) -> APP2(cons, x)
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(app2(app2(app2(insert, f), g), nil), x) -> APP2(app2(cons, x), nil)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(insert, f), g)
APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(max, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(g, x), h)
APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(min, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(g, x)
APP2(ascending_sort, l) -> APP2(app2(app2(sort, min), max), l)
APP2(ascending_sort, l) -> APP2(sort, min)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h))
APP2(ascending_sort, l) -> APP2(app2(sort, min), max)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t))
APP2(descending_sort, l) -> APP2(app2(app2(sort, max), min), l)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(insert, f)

The TRS R consists of the following rules:

app2(app2(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

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(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

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 oriented strictly and are deleted.


APP2(app2(min, app2(s, x)), app2(s, y)) -> APP2(app2(min, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = 2·x1 + x2   
POL(app2(x1, x2)) = 2 + x1 + 2·x2   
POL(min) = 0   
POL(s) = 0   

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(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

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(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

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 oriented strictly and are deleted.


APP2(app2(max, app2(s, x)), app2(s, y)) -> APP2(app2(max, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = 2·x1 + x2   
POL(app2(x1, x2)) = 2 + x1 + 2·x2   
POL(max) = 0   
POL(s) = 0   

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(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

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, h), t)), x) -> APP2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h))
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(f, x)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(app2(sort, f), g), t)
APP2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> APP2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
APP2(ascending_sort, l) -> APP2(app2(app2(sort, min), max), l)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(g, x)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(g, x), h)
APP2(descending_sort, l) -> APP2(app2(app2(sort, max), min), l)
APP2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> APP2(app2(f, x), h)

The TRS R consists of the following rules:

app2(app2(max, 0), x) -> x
app2(app2(max, x), 0) -> x
app2(app2(max, app2(s, x)), app2(s, y)) -> app2(app2(max, x), y)
app2(app2(min, 0), x) -> 0
app2(app2(min, x), 0) -> 0
app2(app2(min, app2(s, x)), app2(s, y)) -> app2(app2(min, x), y)
app2(app2(app2(app2(insert, f), g), nil), x) -> app2(app2(cons, x), nil)
app2(app2(app2(app2(insert, f), g), app2(app2(cons, h), t)), x) -> app2(app2(cons, app2(app2(f, x), h)), app2(app2(app2(app2(insert, f), g), t), app2(app2(g, x), h)))
app2(app2(app2(sort, f), g), nil) -> nil
app2(app2(app2(sort, f), g), app2(app2(cons, h), t)) -> app2(app2(app2(app2(insert, f), g), app2(app2(app2(sort, f), g), t)), h)
app2(ascending_sort, l) -> app2(app2(app2(sort, min), max), l)
app2(descending_sort, l) -> app2(app2(app2(sort, max), min), l)

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