Termination w.r.t. Q proof of /tmp/tmp3YPKnT/Ex7Sorting.trs

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:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(g, x)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(app(app(insert, f), g), z), app(app(g, x), y))
APP(app(min, app(s, x)), app(s, y)) → APP(app(min, x), y)
APP(asort, z) → APP(app(sort, min), max)
APP(app(min, app(s, x)), app(s, y)) → APP(min, x)
APP(dsort, z) → APP(app(sort, max), min)
APP(app(app(app(insert, f), g), nil), y) → APP(app(cons, y), nil)
APP(asort, z) → APP(app(app(sort, min), max), z)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(insert, f), g)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(app(insert, f), g), z)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(insert, f), g), app(app(app(sort, f), g), y))
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(g, x), y)
APP(app(max, app(s, x)), app(s, y)) → APP(max, x)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(insert, f)
APP(app(app(app(insert, f), g), nil), y) → APP(cons, y)
APP(asort, z) → APP(sort, min)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(f, x)
APP(dsort, z) → APP(app(app(sort, max), min), z)
APP(app(max, app(s, x)), app(s, y)) → APP(app(max, x), y)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(sort, f), g), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(cons, app(app(f, x), y))
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(f, x), y)
APP(dsort, z) → APP(sort, max)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(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:

APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(g, x)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(app(app(insert, f), g), z), app(app(g, x), y))
APP(app(min, app(s, x)), app(s, y)) → APP(app(min, x), y)
APP(asort, z) → APP(app(sort, min), max)
APP(app(min, app(s, x)), app(s, y)) → APP(min, x)
APP(dsort, z) → APP(app(sort, max), min)
APP(app(app(app(insert, f), g), nil), y) → APP(app(cons, y), nil)
APP(asort, z) → APP(app(app(sort, min), max), z)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(insert, f), g)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(app(insert, f), g), z)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(insert, f), g), app(app(app(sort, f), g), y))
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(g, x), y)
APP(app(max, app(s, x)), app(s, y)) → APP(max, x)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(insert, f)
APP(app(app(app(insert, f), g), nil), y) → APP(cons, y)
APP(asort, z) → APP(sort, min)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(f, x)
APP(dsort, z) → APP(app(app(sort, max), min), z)
APP(app(max, app(s, x)), app(s, y)) → APP(app(max, x), y)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(sort, f), g), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(cons, app(app(f, x), y))
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(f, x), y)
APP(dsort, z) → APP(sort, max)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

APP(app(min, app(s, x)), app(s, y)) → APP(app(min, x), y)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(app(min, app(s, x)), app(s, y)) → APP(app(min, x), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (15/4)x_2
POL(app(x₁, x₂)) = (1/4)x_1 + (13/4)x_2
POL(s) = 1/4
POL(min) = 0

The value of delta used in the strict ordering is 15/64.
The following usable rules [17] 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:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(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:

APP(app(max, app(s, x)), app(s, y)) → APP(app(max, x), y)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(app(max, app(s, x)), app(s, y)) → APP(app(max, x), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (4)x_2
POL(max) = 0
POL(app(x₁, x₂)) = (4)x_1 + (2)x_2
POL(s) = 4

The value of delta used in the strict ordering is 64.
The following usable rules [17] 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:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(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:

APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(g, x)
APP(dsort, z) → APP(app(app(sort, max), min), z)
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(sort, f), g), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(app(app(insert, f), g), z), app(app(g, x), y))
APP(app(app(sort, f), g), app(app(cons, x), y)) → APP(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(g, x), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(app(f, x), y)
APP(app(app(app(insert, f), g), app(app(cons, x), z)), y) → APP(f, x)
APP(asort, z) → APP(app(app(sort, min), max), z)

The TRS R consists of the following rules:

app(app(app(sort, f), g), nil) → nil
app(app(app(sort, f), g), app(app(cons, x), y)) → app(app(app(app(insert, f), g), app(app(app(sort, f), g), y)), x)
app(app(app(app(insert, f), g), nil), y) → app(app(cons, y), nil)
app(app(app(app(insert, f), g), app(app(cons, x), z)), y) → app(app(cons, app(app(f, x), y)), app(app(app(app(insert, f), g), z), app(app(g, x), y)))
app(app(max, 0), y) → y
app(app(max, x), 0) → x
app(app(max, app(s, x)), app(s, y)) → app(app(max, x), y)
app(app(min, 0), y) → 0
app(app(min, x), 0) → 0
app(app(min, app(s, x)), app(s, y)) → app(app(min, x), y)
app(asort, z) → app(app(app(sort, min), max), z)
app(dsort, z) → app(app(app(sort, max), min), z)

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