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.

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 contains 4 SCCs with 4 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:

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.
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:

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
            ↳ 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:

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.
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:

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
            ↳ 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:

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.
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:

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.
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:

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.
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:

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.