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:

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 oriented strictly and are deleted.

OLD₁(free₁(x)) -> OLD₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(OLD₁(x₁)) = x₁
POL(free₁(x₁)) = 1 + x₁

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 oriented strictly and are deleted.

NEW₁(free₁(x)) -> NEW₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(NEW₁(x₁)) = x₁
POL(free₁(x₁)) = 1 + x₁

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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CHECK₁(x₁)) = 3·x₁
POL(free₁(x₁)) = 3 + 2·x₁
POL(new₁(x₁)) = 3 + x₁
POL(old₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ 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.