Termination w.r.t. Q proof of /tmp/tmpRPLLfN/thiemann35.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:

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), 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:

A(qs, a(a(cons, x), xs)) → A(not, a(le, x))
A(a(append, a(a(cons, x), xs)), ys) → A(a(cons, x), a(a(append, xs), ys))
A(qs, a(a(cons, x), xs)) → A(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))
A(qs, a(a(cons, x), xs)) → A(le, x)
A(qs, a(a(cons, x), xs)) → A(append, a(qs, a(a(filter, a(le, x)), xs)))
A(a(a(if, true), x), xs) → A(cons, x)
A(a(a(if, true), x), xs) → A(a(cons, x), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(a(a(if, a(f, x)), x), a(a(filter, f), xs))
A(qs, a(a(cons, x), xs)) → A(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs)))
A(a(le, a(s, x)), a(s, y)) → A(a(le, x), y)
A(a(append, a(a(cons, x), xs)), ys) → A(a(append, xs), ys)
A(qs, a(a(cons, x), xs)) → A(filter, a(le, x))
A(a(le, a(s, x)), a(s, y)) → A(le, x)
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(append, a(a(cons, x), xs)), ys) → A(append, xs)
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(le, x)), xs))
A(qs, a(a(cons, x), xs)) → A(a(filter, a(le, x)), xs)
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(not, a(le, x))), xs))
A(a(not, f), b) → A(not2, a(f, b))
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(qs, a(a(cons, x), xs)) → A(filter, a(not, a(le, x)))
A(qs, a(a(cons, x), xs)) → A(a(filter, a(not, a(le, x))), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(a(if, a(f, x)), x)
A(a(not, f), b) → A(f, b)
A(a(filter, f), a(a(cons, x), xs)) → A(if, a(f, x))

The TRS R consists of the following rules:

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), 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:

A(qs, a(a(cons, x), xs)) → A(not, a(le, x))
A(a(append, a(a(cons, x), xs)), ys) → A(a(cons, x), a(a(append, xs), ys))
A(qs, a(a(cons, x), xs)) → A(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))
A(qs, a(a(cons, x), xs)) → A(le, x)
A(qs, a(a(cons, x), xs)) → A(append, a(qs, a(a(filter, a(le, x)), xs)))
A(a(a(if, true), x), xs) → A(cons, x)
A(a(a(if, true), x), xs) → A(a(cons, x), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(a(a(if, a(f, x)), x), a(a(filter, f), xs))
A(qs, a(a(cons, x), xs)) → A(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs)))
A(a(le, a(s, x)), a(s, y)) → A(a(le, x), y)
A(a(append, a(a(cons, x), xs)), ys) → A(a(append, xs), ys)
A(qs, a(a(cons, x), xs)) → A(filter, a(le, x))
A(a(le, a(s, x)), a(s, y)) → A(le, x)
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(append, a(a(cons, x), xs)), ys) → A(append, xs)
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(le, x)), xs))
A(qs, a(a(cons, x), xs)) → A(a(filter, a(le, x)), xs)
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(not, a(le, x))), xs))
A(a(not, f), b) → A(not2, a(f, b))
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(qs, a(a(cons, x), xs)) → A(filter, a(not, a(le, x)))
A(qs, a(a(cons, x), xs)) → A(a(filter, a(not, a(le, x))), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(a(if, a(f, x)), x)
A(a(not, f), b) → A(f, b)
A(a(filter, f), a(a(cons, x), xs)) → A(if, a(f, x))

The TRS R consists of the following rules:

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), 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 16 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

A(a(le, a(s, x)), a(s, y)) → A(a(le, x), y)

The TRS R consists of the following rules:

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), 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.

A(a(le, a(s, x)), a(s, y)) → A(a(le, x), y)

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

POL(a(x₁, x₂)) = 4 + (2)x_2
POL(A(x₁, x₂)) = (4)x_1 + (3)x_2
POL(s) = 0
POL(le) = 0

The value of delta used in the strict ordering is 44.
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:

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), 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:

A(a(append, a(a(cons, x), xs)), ys) → A(a(append, xs), ys)

The TRS R consists of the following rules:

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), 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.

A(a(append, a(a(cons, x), xs)), ys) → A(a(append, xs), ys)

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

POL(append) = 0
POL(cons) = 2
POL(a(x₁, x₂)) = 1 + (3)x_2
POL(A(x₁, x₂)) = (4)x_1

The value of delta used in the strict ordering is 12.
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:

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), 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

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

A(qs, a(a(cons, x), xs)) → A(a(filter, a(not, a(le, x))), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(qs, a(a(cons, x), xs)) → A(a(filter, a(le, x)), xs)
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(le, x)), xs))
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(not, a(le, x))), xs))
A(a(not, f), b) → A(f, b)
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)

The TRS R consists of the following rules:

a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))

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