Termination w.r.t. Q proof of /tmp/tmpir5PZV/TreeLevels.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(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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))

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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(combine, xs), app(app(cons, ys), yss)) → APP(zip, xs)
APP(levels, app(app(node, x), xs)) → APP(map, levels)
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(levels, app(app(node, x), xs)) → APP(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(levels, app(app(node, x), xs)) → APP(app(map, levels), xs)
APP(levels, app(app(node, x), xs)) → APP(cons, app(app(cons, x), nil))
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(zip, xs), ys)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(cons, app(app(append, xs), ys))
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(zip, xss), yss)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(append, xs), ys)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(combine, app(app(zip, xs), ys))
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(append, xs)
APP(levels, app(app(node, x), xs)) → APP(cons, x)
APP(levels, app(app(node, x), xs)) → APP(app(combine, nil), app(app(map, levels), xs))
APP(levels, app(app(node, x), xs)) → APP(combine, nil)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(combine, app(app(zip, xs), ys)), yss)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(zip, xss)
APP(levels, app(app(node, x), xs)) → APP(app(cons, x), nil)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), 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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(combine, xs), app(app(cons, ys), yss)) → APP(zip, xs)
APP(levels, app(app(node, x), xs)) → APP(map, levels)
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(levels, app(app(node, x), xs)) → APP(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(levels, app(app(node, x), xs)) → APP(app(map, levels), xs)
APP(levels, app(app(node, x), xs)) → APP(cons, app(app(cons, x), nil))
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(zip, xs), ys)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(cons, app(app(append, xs), ys))
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(zip, xss), yss)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(append, xs), ys)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(combine, app(app(zip, xs), ys))
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(append, xs)
APP(levels, app(app(node, x), xs)) → APP(cons, x)
APP(levels, app(app(node, x), xs)) → APP(app(combine, nil), app(app(map, levels), xs))
APP(levels, app(app(node, x), xs)) → APP(combine, nil)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(combine, app(app(zip, xs), ys)), yss)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(zip, xss)
APP(levels, app(app(node, x), xs)) → APP(app(cons, x), nil)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), 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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))

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 19 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ 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(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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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

          ↳ 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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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

          ↳ QDP

          ↳ QDP

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

APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(zip, xss), yss)

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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(zip, xss), yss)

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

POL(APP(x₁, x₂)) = (3/2)x_1 + (15/4)x_2
POL(cons) = 4
POL(zip) = 0
POL(app(x₁, x₂)) = 3/4 + (13/4)x_2

The value of delta used in the strict ordering is 207/32.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ 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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(combine, app(app(zip, xs), ys)), yss)

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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(combine, xs), app(app(cons, ys), yss)) → APP(app(combine, app(app(zip, xs), ys)), yss)

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

POL(APP(x₁, x₂)) = (4)x_2
POL(append) = 13/4
POL(cons) = 4
POL(zip) = 1/4
POL(combine) = 0
POL(app(x₁, x₂)) = (4)x_1 + (2)x_2
POL(nil) = 0

The value of delta used in the strict ordering is 256.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ 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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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

          ↳ 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(levels, app(app(node, x), xs)) → APP(app(map, levels), xs)
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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(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.

APP(levels, app(app(node, x), xs)) → APP(app(map, levels), xs)

Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = x_1 + (1/4)x_2
POL(cons) = 3/4
POL(levels) = 0
POL(map) = 1/4
POL(node) = 4
POL(app(x₁, x₂)) = (1/4)x_1 + (4)x_2

The value of delta used in the strict ordering is 3/256.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

APP(levels, app(app(node, x), xs)) → APP(app(map, levels), 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(append, xs), nil) → xs
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(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.