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 Order [17,21] with Interpretation:

POL( free1(x1) ) = 3x1 + 3


POL( OLD1(x1) ) = 2x1 + 3



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 Order [17,21] with Interpretation:

POL( free1(x1) ) = 3x1 + 3


POL( NEW1(x1) ) = 2x1 + 3



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 Order [17,21] with Interpretation:

POL( free1(x1) ) = 2x1 + 3


POL( new1(x1) ) = 2x1 + 3


POL( old1(x1) ) = 3x1 + 2


POL( CHECK1(x1) ) = 3x1



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.