Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/SK90SLASH4.28.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:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.

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

MAX₁(g₂(g₂(g₂(x, y), z), u)) -> MAX₁(g₂(g₂(x, y), z))
MEM₂(g₂(x, y), z) -> MEM₂(x, z)
F₂(x, g₂(y, z)) -> F₂(x, y)
++¹₂(x, g₂(y, z)) -> ++¹₂(x, y)
MEM₂(x, max₁(x)) -> NULL₁(x)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

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:

MAX₁(g₂(g₂(g₂(x, y), z), u)) -> MAX₁(g₂(g₂(x, y), z))
MEM₂(g₂(x, y), z) -> MEM₂(x, z)
F₂(x, g₂(y, z)) -> F₂(x, y)
++¹₂(x, g₂(y, z)) -> ++¹₂(x, y)
MEM₂(x, max₁(x)) -> NULL₁(x)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MAX₁(g₂(g₂(g₂(x, y), z), u)) -> MAX₁(g₂(g₂(x, y), z))

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MAX₁(g₂(g₂(g₂(x, y), z), u)) -> MAX₁(g₂(g₂(x, y), z))

Used argument filtering: MAX₁(x1) = x1
g₂(x1, x2) = g₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

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

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

MEM₂(g₂(x, y), z) -> MEM₂(x, z)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MEM₂(g₂(x, y), z) -> MEM₂(x, z)

Used argument filtering: MEM₂(x1, x2) = x1
g₂(x1, x2) = g₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

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

            ↳ QDPAfsSolverProof

          ↳ QDP

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

++¹₂(x, g₂(y, z)) -> ++¹₂(x, y)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

++¹₂(x, g₂(y, z)) -> ++¹₂(x, y)

Used argument filtering: ++¹₂(x1, x2) = x2
g₂(x1, x2) = g₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

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

            ↳ QDPAfsSolverProof

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

F₂(x, g₂(y, z)) -> F₂(x, y)

The TRS R consists of the following rules:

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₂(x, g₂(y, z)) -> F₂(x, y)

Used argument filtering: F₂(x1, x2) = x2
g₂(x1, x2) = g₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

f₂(x, nil) -> g₂(nil, x)
f₂(x, g₂(y, z)) -> g₂(f₂(x, y), z)
++₂(x, nil) -> x
++₂(x, g₂(y, z)) -> g₂(++₂(x, y), z)
null₁(nil) -> true
null₁(g₂(x, y)) -> false
mem₂(nil, y) -> false
mem₂(g₂(x, y), z) -> or₂(=₂(y, z), mem₂(x, z))
mem₂(x, max₁(x)) -> not₁(null₁(x))
max₁(g₂(g₂(nil, x), y)) -> max'₂(x, y)
max₁(g₂(g₂(g₂(x, y), z), u)) -> max'₂(max₁(g₂(g₂(x, y), z)), u)

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.