Termination w.r.t. Q proof of TRS/AProVELiveness6.2.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(sent₁(x)) -> top₁(check₁(rest₁(x)))
rest₁(nil) -> sent₁(nil)
rest₁(cons₂(x, y)) -> sent₁(y)
check₁(sent₁(x)) -> sent₁(check₁(x))
check₁(rest₁(x)) -> rest₁(check₁(x))
check₁(cons₂(x, y)) -> cons₂(check₁(x), y)
check₁(cons₂(x, y)) -> cons₂(x, check₁(y))
check₁(cons₂(x, y)) -> cons₂(x, y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

top₁(sent₁(x)) -> top₁(check₁(rest₁(x)))
rest₁(nil) -> sent₁(nil)
rest₁(cons₂(x, y)) -> sent₁(y)
check₁(sent₁(x)) -> sent₁(check₁(x))
check₁(rest₁(x)) -> rest₁(check₁(x))
check₁(cons₂(x, y)) -> cons₂(check₁(x), y)
check₁(cons₂(x, y)) -> cons₂(x, check₁(y))
check₁(cons₂(x, y)) -> cons₂(x, y)

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:

CHECK₁(cons₂(x, y)) -> CHECK₁(y)
TOP₁(sent₁(x)) -> REST₁(x)
TOP₁(sent₁(x)) -> CHECK₁(rest₁(x))
TOP₁(sent₁(x)) -> TOP₁(check₁(rest₁(x)))
CHECK₁(rest₁(x)) -> REST₁(check₁(x))
CHECK₁(rest₁(x)) -> CHECK₁(x)
CHECK₁(cons₂(x, y)) -> CHECK₁(x)
CHECK₁(sent₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top₁(sent₁(x)) -> top₁(check₁(rest₁(x)))
rest₁(nil) -> sent₁(nil)
rest₁(cons₂(x, y)) -> sent₁(y)
check₁(sent₁(x)) -> sent₁(check₁(x))
check₁(rest₁(x)) -> rest₁(check₁(x))
check₁(cons₂(x, y)) -> cons₂(check₁(x), y)
check₁(cons₂(x, y)) -> cons₂(x, check₁(y))
check₁(cons₂(x, y)) -> cons₂(x, y)

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:

CHECK₁(cons₂(x, y)) -> CHECK₁(y)
TOP₁(sent₁(x)) -> REST₁(x)
TOP₁(sent₁(x)) -> CHECK₁(rest₁(x))
TOP₁(sent₁(x)) -> TOP₁(check₁(rest₁(x)))
CHECK₁(rest₁(x)) -> REST₁(check₁(x))
CHECK₁(rest₁(x)) -> CHECK₁(x)
CHECK₁(cons₂(x, y)) -> CHECK₁(x)
CHECK₁(sent₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top₁(sent₁(x)) -> top₁(check₁(rest₁(x)))
rest₁(nil) -> sent₁(nil)
rest₁(cons₂(x, y)) -> sent₁(y)
check₁(sent₁(x)) -> sent₁(check₁(x))
check₁(rest₁(x)) -> rest₁(check₁(x))
check₁(cons₂(x, y)) -> cons₂(check₁(x), y)
check₁(cons₂(x, y)) -> cons₂(x, check₁(y))
check₁(cons₂(x, y)) -> cons₂(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

CHECK₁(cons₂(x, y)) -> CHECK₁(y)
CHECK₁(rest₁(x)) -> CHECK₁(x)
CHECK₁(cons₂(x, y)) -> CHECK₁(x)
CHECK₁(sent₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top₁(sent₁(x)) -> top₁(check₁(rest₁(x)))
rest₁(nil) -> sent₁(nil)
rest₁(cons₂(x, y)) -> sent₁(y)
check₁(sent₁(x)) -> sent₁(check₁(x))
check₁(rest₁(x)) -> rest₁(check₁(x))
check₁(cons₂(x, y)) -> cons₂(check₁(x), y)
check₁(cons₂(x, y)) -> cons₂(x, check₁(y))
check₁(cons₂(x, y)) -> cons₂(x, y)

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(cons₂(x, y)) -> CHECK₁(y)
CHECK₁(cons₂(x, y)) -> CHECK₁(x)
CHECK₁(sent₁(x)) -> CHECK₁(x)

The remaining pairs can at least be oriented weakly.

CHECK₁(rest₁(x)) -> CHECK₁(x)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( rest₁(x₁) ) = x₁

POL( sent₁(x₁) ) = 2x₁ + 3

POL( cons₂(x₁, x₂) ) = 2x₁ + x₂ + 1

POL( CHECK₁(x₁) ) = 3x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

CHECK₁(rest₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top₁(sent₁(x)) -> top₁(check₁(rest₁(x)))
rest₁(nil) -> sent₁(nil)
rest₁(cons₂(x, y)) -> sent₁(y)
check₁(sent₁(x)) -> sent₁(check₁(x))
check₁(rest₁(x)) -> rest₁(check₁(x))
check₁(cons₂(x, y)) -> cons₂(check₁(x), y)
check₁(cons₂(x, y)) -> cons₂(x, check₁(y))
check₁(cons₂(x, y)) -> cons₂(x, y)

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(rest₁(x)) -> CHECK₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( rest₁(x₁) ) = 3x₁ + 3

POL( CHECK₁(x₁) ) = 2x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

top₁(sent₁(x)) -> top₁(check₁(rest₁(x)))
rest₁(nil) -> sent₁(nil)
rest₁(cons₂(x, y)) -> sent₁(y)
check₁(sent₁(x)) -> sent₁(check₁(x))
check₁(rest₁(x)) -> rest₁(check₁(x))
check₁(cons₂(x, y)) -> cons₂(check₁(x), y)
check₁(cons₂(x, y)) -> cons₂(x, check₁(y))
check₁(cons₂(x, y)) -> cons₂(x, y)

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

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

TOP₁(sent₁(x)) -> TOP₁(check₁(rest₁(x)))

The TRS R consists of the following rules:

top₁(sent₁(x)) -> top₁(check₁(rest₁(x)))
rest₁(nil) -> sent₁(nil)
rest₁(cons₂(x, y)) -> sent₁(y)
check₁(sent₁(x)) -> sent₁(check₁(x))
check₁(rest₁(x)) -> rest₁(check₁(x))
check₁(cons₂(x, y)) -> cons₂(check₁(x), y)
check₁(cons₂(x, y)) -> cons₂(x, check₁(y))
check₁(cons₂(x, y)) -> cons₂(x, y)

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