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.

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 contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHECK1(cons2(x, y)) -> CHECK1(y)
CHECK1(cons2(x, y)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
rest1(x1)  =  x1
sent1(x1)  =  x1
Used ordering: Quasi Precedence: trivial 


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

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

CHECK1(rest1(x)) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHECK1(sent1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
rest1(x1)  =  x1
sent1(x1)  =  sent1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHECK1(rest1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
rest1(x1)  =  rest1(x1)
Used ordering: Quasi Precedence: trivial 


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