Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/AProVESLASHLiveness6.1.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:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

OLD₁(free₁(x)) -> OLD₁(x)
CHECK₁(new₁(x)) -> CHECK₁(x)
TOP₁(free₁(x)) -> TOP₁(check₁(new₁(x)))
NEW₁(free₁(x)) -> NEW₁(x)
CHECK₁(old₁(x)) -> CHECK₁(x)
CHECK₁(new₁(x)) -> NEW₁(check₁(x))
TOP₁(free₁(x)) -> CHECK₁(new₁(x))
CHECK₁(free₁(x)) -> CHECK₁(x)
TOP₁(free₁(x)) -> NEW₁(x)
CHECK₁(old₁(x)) -> OLD₁(check₁(x))

The TRS R consists of the following rules:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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:

OLD₁(free₁(x)) -> OLD₁(x)
CHECK₁(new₁(x)) -> CHECK₁(x)
TOP₁(free₁(x)) -> TOP₁(check₁(new₁(x)))
NEW₁(free₁(x)) -> NEW₁(x)
CHECK₁(old₁(x)) -> CHECK₁(x)
CHECK₁(new₁(x)) -> NEW₁(check₁(x))
TOP₁(free₁(x)) -> CHECK₁(new₁(x))
CHECK₁(free₁(x)) -> CHECK₁(x)
TOP₁(free₁(x)) -> NEW₁(x)
CHECK₁(old₁(x)) -> OLD₁(check₁(x))

The TRS R consists of the following rules:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

OLD₁(free₁(x)) -> OLD₁(x)

The TRS R consists of the following rules:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

OLD₁(free₁(x)) -> OLD₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
OLD₁(x1) = OLD₁(x1)
free₁(x1) = free₁(x1)

Lexicographic Path Order [19].
Precedence:

free₁ > OLD₁

The following usable rules [14] 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:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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:

NEW₁(free₁(x)) -> NEW₁(x)

The TRS R consists of the following rules:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

NEW₁(free₁(x)) -> NEW₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
NEW₁(x1) = NEW₁(x1)
free₁(x1) = free₁(x1)

Lexicographic Path Order [19].
Precedence:

free₁ > NEW₁

The following usable rules [14] 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:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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:

CHECK₁(new₁(x)) -> CHECK₁(x)
CHECK₁(old₁(x)) -> CHECK₁(x)
CHECK₁(free₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

CHECK₁(old₁(x)) -> CHECK₁(x)

The remaining pairs can at least by weakly be oriented.

CHECK₁(new₁(x)) -> CHECK₁(x)
CHECK₁(free₁(x)) -> CHECK₁(x)

Used ordering: Combined order from the following AFS and order.
CHECK₁(x1) = CHECK₁(x1)
new₁(x1) = x1
old₁(x1) = old₁(x1)
free₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

[CHECK₁, old_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

CHECK₁(new₁(x)) -> CHECK₁(x)
CHECK₁(free₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

CHECK₁(free₁(x)) -> CHECK₁(x)

The remaining pairs can at least by weakly be oriented.

CHECK₁(new₁(x)) -> CHECK₁(x)

Used ordering: Combined order from the following AFS and order.
CHECK₁(x1) = CHECK₁(x1)
new₁(x1) = x1
free₁(x1) = free₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

CHECK₁(new₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

CHECK₁(new₁(x)) -> CHECK₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
CHECK₁(x1) = CHECK₁(x1)
new₁(x1) = new₁(x1)

Lexicographic Path Order [19].
Precedence:

new₁ > CHECK₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

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

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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

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

TOP₁(free₁(x)) -> TOP₁(check₁(new₁(x)))

The TRS R consists of the following rules:

top₁(free₁(x)) -> top₁(check₁(new₁(x)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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