Termination w.r.t. Q proof of /tmp/tmpIvkjsv/013.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(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

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(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(app(flatwith, f), app(leaf, x)) → APP(app(cons, app(f, x)), nil)
APP(app(flatwith, f), app(leaf, x)) → APP(cons, app(f, x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(append, app(app(flatwith, f), x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(flatwith, f)
APP(app(flatwith, f), app(node, xs)) → APP(flatwithsub, f)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

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(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(app(flatwith, f), app(leaf, x)) → APP(app(cons, app(f, x)), nil)
APP(app(flatwith, f), app(leaf, x)) → APP(cons, app(f, x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(append, app(app(flatwith, f), x))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(flatwith, f)
APP(app(flatwith, f), app(node, xs)) → APP(flatwithsub, f)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

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(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)

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

POL(APP(x₁, x₂)) = (3/2)x_1
POL(append) = 0
POL(cons) = 9/4
POL(app(x₁, x₂)) = (1/2)x_1 + (9/4)x_2

The value of delta used in the strict ordering is 243/128.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

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

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

APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwith, f), app(leaf, x)) → APP(f, x)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

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(flatwith, f), app(leaf, x)) → APP(f, x)

The remaining pairs can at least be oriented weakly.

APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)

Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = x_1
POL(flatwithsub) = 7/2
POL(cons) = 0
POL(node) = 1/4
POL(leaf) = 11/4
POL(app(x₁, x₂)) = 2 + (4)x_2
POL(flatwith) = 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

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

APP(app(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)

The TRS R consists of the following rules:

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

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(flatwith, f), app(node, xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwithsub, f), xs)
APP(app(flatwithsub, f), app(app(cons, x), xs)) → APP(app(flatwith, f), x)

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

POL(APP(x₁, x₂)) = (4)x_2
POL(flatwithsub) = 3/2
POL(cons) = 1/4
POL(node) = 1
POL(app(x₁, x₂)) = 3/2 + (1/2)x_1 + (4)x_2
POL(flatwith) = 5/2

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(flatwith, f), app(leaf, x)) → app(app(cons, app(f, x)), nil)
app(app(flatwith, f), app(node, xs)) → app(app(flatwithsub, f), xs)
app(app(flatwithsub, f), nil) → nil
app(app(flatwithsub, f), app(app(cons, x), xs)) → app(app(append, app(app(flatwith, f), x)), app(app(flatwithsub, f), xs))

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.