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:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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.

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

TOP22(x, free1(y)) -> TOP12(x, check1(new1(y)))
TOP22(x, free1(y)) -> CHECK1(new1(y))
TOP12(free1(x), y) -> CHECK1(new1(x))
TOP22(x, free1(y)) -> TOP12(new1(x), check1(y))
TOP22(x, free1(y)) -> TOP12(check1(x), new1(y))
TOP22(x, free1(y)) -> CHECK1(x)
TOP22(x, free1(y)) -> TOP12(check1(new1(x)), y)
CHECK1(free1(x)) -> CHECK1(x)
CHECK1(old1(x)) -> OLD1(check1(x))
TOP12(free1(x), y) -> CHECK1(new1(y))
TOP12(free1(x), y) -> CHECK1(x)
TOP12(free1(x), y) -> NEW1(x)
CHECK1(new1(x)) -> CHECK1(x)
TOP22(x, free1(y)) -> NEW1(x)
TOP12(free1(x), y) -> TOP22(check1(x), new1(y))
TOP12(free1(x), y) -> TOP22(new1(x), check1(y))
TOP22(x, free1(y)) -> CHECK1(new1(x))
TOP12(free1(x), y) -> NEW1(y)
TOP12(free1(x), y) -> CHECK1(y)
TOP12(free1(x), y) -> TOP22(check1(new1(x)), y)
OLD1(free1(x)) -> OLD1(x)
CHECK1(old1(x)) -> CHECK1(x)
CHECK1(new1(x)) -> NEW1(check1(x))
TOP22(x, free1(y)) -> NEW1(y)
TOP22(x, free1(y)) -> CHECK1(y)
TOP12(free1(x), y) -> TOP22(x, check1(new1(y)))
NEW1(free1(x)) -> NEW1(x)

The TRS R consists of the following rules:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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:

TOP22(x, free1(y)) -> TOP12(x, check1(new1(y)))
TOP22(x, free1(y)) -> CHECK1(new1(y))
TOP12(free1(x), y) -> CHECK1(new1(x))
TOP22(x, free1(y)) -> TOP12(new1(x), check1(y))
TOP22(x, free1(y)) -> TOP12(check1(x), new1(y))
TOP22(x, free1(y)) -> CHECK1(x)
TOP22(x, free1(y)) -> TOP12(check1(new1(x)), y)
CHECK1(free1(x)) -> CHECK1(x)
CHECK1(old1(x)) -> OLD1(check1(x))
TOP12(free1(x), y) -> CHECK1(new1(y))
TOP12(free1(x), y) -> CHECK1(x)
TOP12(free1(x), y) -> NEW1(x)
CHECK1(new1(x)) -> CHECK1(x)
TOP22(x, free1(y)) -> NEW1(x)
TOP12(free1(x), y) -> TOP22(check1(x), new1(y))
TOP12(free1(x), y) -> TOP22(new1(x), check1(y))
TOP22(x, free1(y)) -> CHECK1(new1(x))
TOP12(free1(x), y) -> NEW1(y)
TOP12(free1(x), y) -> CHECK1(y)
TOP12(free1(x), y) -> TOP22(check1(new1(x)), y)
OLD1(free1(x)) -> OLD1(x)
CHECK1(old1(x)) -> CHECK1(x)
CHECK1(new1(x)) -> NEW1(check1(x))
TOP22(x, free1(y)) -> NEW1(y)
TOP22(x, free1(y)) -> CHECK1(y)
TOP12(free1(x), y) -> TOP22(x, check1(new1(y)))
NEW1(free1(x)) -> NEW1(x)

The TRS R consists of the following rules:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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 contains 4 SCCs with 14 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

OLD1(free1(x)) -> OLD1(x)
Used argument filtering: OLD1(x1)  =  x1
free1(x1)  =  free1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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
            ↳ QDPAfsSolverProof
          ↳ 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:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

NEW1(free1(x)) -> NEW1(x)
Used argument filtering: NEW1(x1)  =  x1
free1(x1)  =  free1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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
            ↳ QDPAfsSolverProof
          ↳ 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:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHECK1(free1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
new1(x1)  =  x1
old1(x1)  =  x1
free1(x1)  =  free1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHECK1(old1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
new1(x1)  =  x1
old1(x1)  =  old1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ 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:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHECK1(new1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
new1(x1)  =  new1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP

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

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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:

TOP22(x, free1(y)) -> TOP12(x, check1(new1(y)))
TOP12(free1(x), y) -> TOP22(check1(new1(x)), y)
TOP12(free1(x), y) -> TOP22(x, check1(new1(y)))
TOP12(free1(x), y) -> TOP22(check1(x), new1(y))
TOP12(free1(x), y) -> TOP22(new1(x), check1(y))
TOP22(x, free1(y)) -> TOP12(check1(x), new1(y))
TOP22(x, free1(y)) -> TOP12(new1(x), check1(y))
TOP22(x, free1(y)) -> TOP12(check1(new1(x)), y)

The TRS R consists of the following rules:

top12(free1(x), y) -> top22(check1(new1(x)), y)
top12(free1(x), y) -> top22(new1(x), check1(y))
top12(free1(x), y) -> top22(check1(x), new1(y))
top12(free1(x), y) -> top22(x, check1(new1(y)))
top22(x, free1(y)) -> top12(check1(new1(x)), y)
top22(x, free1(y)) -> top12(new1(x), check1(y))
top22(x, free1(y)) -> top12(check1(x), new1(y))
top22(x, free1(y)) -> top12(x, check1(new1(y)))
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.