Termination w.r.t. Q proof of /tmp/tmphoOcwj/TreeHeight.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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(height, app(app(node, x), xs)) → APP(s, app(app(maxlist, 0), app(app(map, height), xs)))
APP(height, app(app(node, x), xs)) → APP(app(maxlist, 0), app(app(map, height), xs))
APP(height, app(app(node, x), xs)) → APP(app(map, height), xs)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(if, app(app(le, x), y))
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(le, x)
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
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)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(maxlist, y), ys)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(maxlist, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(le, x), y)
APP(height, app(app(node, x), xs)) → APP(maxlist, 0)
APP(height, app(app(node, x), xs)) → APP(map, height)
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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(height, app(app(node, x), xs)) → APP(s, app(app(maxlist, 0), app(app(map, height), xs)))
APP(height, app(app(node, x), xs)) → APP(app(maxlist, 0), app(app(map, height), xs))
APP(height, app(app(node, x), xs)) → APP(app(map, height), xs)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(if, app(app(le, x), y))
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(le, x)
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
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)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(maxlist, y), ys)
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(maxlist, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(le, x), y)
APP(height, app(app(node, x), xs)) → APP(maxlist, 0)
APP(height, app(app(node, x), xs)) → APP(map, height)
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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), xs)))

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 12 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), 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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

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

POL(APP(x₁, x₂)) = (1/4)x_1
POL(app(x₁, x₂)) = 1 + (2)x_2
POL(s) = 0
POL(le) = 0

The value of delta used in the strict ordering is 1/2.
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(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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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

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

APP(app(maxlist, x), app(app(cons, y), ys)) → APP(app(maxlist, y), 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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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(maxlist, x), app(app(cons, y), ys)) → APP(app(maxlist, y), ys)

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

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

The value of delta used in the strict ordering is 1/8.
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(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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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

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

APP(height, app(app(node, x), xs)) → APP(app(map, height), xs)
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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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(height, app(app(node, x), xs)) → APP(app(map, height), xs)

Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (1/4)x_2
POL(height) = 0
POL(cons) = 2
POL(map) = 0
POL(node) = 0
POL(app(x₁, x₂)) = (1/4)x_1 + (4)x_2

The value of delta used in the strict ordering is 1/32.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

APP(height, app(app(node, x), xs)) → APP(app(map, height), 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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(app(maxlist, x), app(app(cons, y), ys)) → app(app(if, app(app(le, x), y)), app(app(maxlist, y), ys))
app(app(maxlist, x), nil) → x
app(height, app(app(node, x), xs)) → app(s, app(app(maxlist, 0), app(app(map, height), 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.