Termination w.r.t. Q proof of TRS/AProVELiveness6.1.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:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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:

OLD(free(x)) → OLD(x)
CHECK(new(x)) → CHECK(x)
TOP(free(x)) → TOP(check(new(x)))
NEW(free(x)) → NEW(x)
CHECK(old(x)) → CHECK(x)
CHECK(new(x)) → NEW(check(x))
TOP(free(x)) → CHECK(new(x))
CHECK(free(x)) → CHECK(x)
TOP(free(x)) → NEW(x)
CHECK(old(x)) → OLD(check(x))

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

OLD(free(x)) → OLD(x)
CHECK(new(x)) → CHECK(x)
TOP(free(x)) → TOP(check(new(x)))
NEW(free(x)) → NEW(x)
CHECK(old(x)) → CHECK(x)
CHECK(new(x)) → NEW(check(x))
TOP(free(x)) → CHECK(new(x))
CHECK(free(x)) → CHECK(x)
TOP(free(x)) → NEW(x)
CHECK(old(x)) → OLD(check(x))

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

OLD(free(x)) → OLD(x)
CHECK(new(x)) → CHECK(x)
TOP(free(x)) → TOP(check(new(x)))
NEW(free(x)) → NEW(x)
CHECK(new(x)) → NEW(check(x))
CHECK(old(x)) → CHECK(x)
TOP(free(x)) → CHECK(new(x))
CHECK(free(x)) → CHECK(x)
TOP(free(x)) → NEW(x)
CHECK(old(x)) → OLD(check(x))

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

OLD(free(x)) → OLD(x)

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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 oriented strictly and are deleted.

OLD(free(x)) → OLD(x)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

NEW(free(x)) → NEW(x)

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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 oriented strictly and are deleted.

NEW(free(x)) → NEW(x)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

CHECK(new(x)) → CHECK(x)
CHECK(old(x)) → CHECK(x)
CHECK(free(x)) → CHECK(x)

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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 oriented strictly and are deleted.

CHECK(new(x)) → CHECK(x)

The remaining pairs can at least be oriented weakly.

CHECK(old(x)) → CHECK(x)
CHECK(free(x)) → CHECK(x)

Used ordering: Combined order from the following AFS and order.
CHECK(x1) = CHECK(x1)
new(x1) = new(x1)
old(x1) = x1
free(x1) = x1

Recursive Path Order [2].
Precedence:

new₁ > CHECK₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

CHECK(old(x)) → CHECK(x)
CHECK(free(x)) → CHECK(x)

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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 oriented strictly and are deleted.

CHECK(old(x)) → CHECK(x)

The remaining pairs can at least be oriented weakly.

CHECK(free(x)) → CHECK(x)

Used ordering: Combined order from the following AFS and order.
CHECK(x1) = CHECK(x1)
old(x1) = old(x1)
free(x1) = x1

Recursive Path Order [2].
Precedence:

old₁ > CHECK₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ QDP

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

CHECK(free(x)) → CHECK(x)

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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 oriented strictly and are deleted.

CHECK(free(x)) → CHECK(x)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

TOP(free(x)) → TOP(check(new(x)))

The TRS R consists of the following rules:

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(x)

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