Termination w.r.t. Q proof of /tmp/tmpP6c6E1/kabasci06.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(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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(id, app(p, x)) → APP(s, app(id, x))
APP(app(div, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, app(p, app(s, x)))
APP(id, x) → APP(s, x)
APP(app(div, app(s, x)), app(s, y)) → APP(id, y)
APP(id, x) → APP(s, app(s, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(id, x) → APP(s, app(s, app(s, x)))
APP(id, app(p, x)) → APP(id, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(p, app(s, x))
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), app(id, y))), app(s, y))
APP(app(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), app(id, y)))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), app(id, y))
APP(id, app(p, x)) → APP(id, app(s, app(id, x)))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(minus, app(s, x)), app(s, y)) → APP(p, app(s, y))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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(id, app(p, x)) → APP(s, app(id, x))
APP(app(div, app(s, x)), app(s, y)) → APP(minus, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(minus, app(p, app(s, x)))
APP(id, x) → APP(s, x)
APP(app(div, app(s, x)), app(s, y)) → APP(id, y)
APP(id, x) → APP(s, app(s, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(div, app(s, x)), app(s, y)) → APP(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(id, x) → APP(s, app(s, app(s, x)))
APP(id, app(p, x)) → APP(id, x)
APP(app(minus, app(s, x)), app(s, y)) → APP(p, app(s, x))
APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), app(id, y))), app(s, y))
APP(app(div, app(s, x)), app(s, y)) → APP(div, app(app(minus, x), app(id, y)))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))
APP(app(div, app(s, x)), app(s, y)) → APP(app(minus, x), app(id, y))
APP(id, app(p, x)) → APP(id, app(s, app(id, x)))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(minus, app(s, x)), app(s, y)) → APP(p, app(s, y))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 15 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP(id, app(p, x)) → APP(id, x)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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(id, app(p, x)) → APP(id, x)

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

POL(APP(x₁, x₂)) = (2)x_2
POL(p) = 3
POL(app(x₁, x₂)) = 1 + (4)x_2
POL(id) = 0

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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

          ↳ QDP

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

APP(app(minus, app(s, x)), app(s, y)) → APP(app(minus, app(p, app(s, x))), app(p, app(s, y)))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP(app(div, app(s, x)), app(s, y)) → APP(app(div, app(app(minus, x), app(id, y))), app(s, y))

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

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

POL(APP(x₁, x₂)) = (4)x_1 + (3)x_2
POL(cons) = 1
POL(map) = 4
POL(app(x₁, x₂)) = 4 + x_1 + (2)x_2

The value of delta used in the strict ordering is 27.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(minus, x), 0) → x
app(app(minus, app(s, x)), app(s, y)) → app(app(minus, app(p, app(s, x))), app(p, app(s, y)))
app(p, app(s, x)) → x
app(app(div, 0), app(s, y)) → 0
app(app(div, app(s, x)), app(s, y)) → app(s, app(app(div, app(app(minus, x), app(id, y))), app(s, y)))
app(id, x) → x
app(id, x) → app(s, app(s, app(s, x)))
app(id, app(p, x)) → app(id, app(s, app(id, x)))

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.