Termination w.r.t. Q proof of /tmp/tmphNSNcl/Ex8Polymorphic.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), 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(app(twice, f), x) → APP(f, app(f, x))
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(app(fmap, t_f), x)
APP(app(twice, f), x) → APP(f, x)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(app(cons, app(f, x)), app(app(fmap, t_f), x))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(cons, app(f, x))
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(f, x)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(fmap, t_f)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))

The TRS R consists of the following rules:

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), 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(app(twice, f), x) → APP(f, app(f, x))
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(app(fmap, t_f), x)
APP(app(twice, f), x) → APP(f, x)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(app(cons, app(f, x)), app(app(fmap, t_f), x))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(cons, app(f, x))
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(f, x)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(fmap, t_f)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))

The TRS R consists of the following rules:

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

APP(app(twice, f), x) → APP(f, app(f, x))
APP(app(twice, f), x) → APP(f, x)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(f, x)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The TRS R consists of the following rules:

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), 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(twice, f), x) → APP(f, app(f, x))
APP(app(twice, f), x) → APP(f, x)
APP(app(fmap, app(app(cons, f), t_f)), x) → APP(f, x)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)

The remaining pairs can at least be oriented weakly.

APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (1/4)x_1
POL(t_f) = 0
POL(cons) = 0
POL(map) = 1/4
POL(app(x₁, x₂)) = (1/4)x_1 + (11/4)x_2
POL(twice) = 1
POL(fmap) = 1/2
POL(nil) = 7/4

The value of delta used in the strict ordering is 1/64.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The TRS R consists of the following rules:

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), 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, h), t)) → APP(app(map, f), t)

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

POL(APP(x₁, x₂)) = (4)x_2
POL(cons) = 2
POL(map) = 0
POL(app(x₁, x₂)) = 1 + (4)x_2

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

app(app(twice, f), x) → app(f, app(f, x))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(fmap, nil), x) → nil
app(app(fmap, app(app(cons, f), t_f)), x) → app(app(cons, app(f, x)), app(app(fmap, t_f), 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.