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:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)

Q is empty.

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

TOP1(rec1(up1(x))) -> REC1(x)
CHECK1(no1(x)) -> CHECK1(x)
REC1(bot) -> SENT1(bot)
REC1(rec1(x)) -> SENT1(rec1(x))
TOP1(sent1(up1(x))) -> CHECK1(rec1(x))
CHECK1(rec1(x)) -> CHECK1(x)
REC1(up1(x)) -> REC1(x)
TOP1(no1(up1(x))) -> REC1(x)
TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(no1(x)) -> SENT1(rec1(x))
CHECK1(sent1(x)) -> SENT1(check1(x))
CHECK1(rec1(x)) -> REC1(check1(x))
TOP1(no1(up1(x))) -> CHECK1(rec1(x))
CHECK1(no1(x)) -> NO1(check1(x))
REC1(sent1(x)) -> SENT1(rec1(x))
CHECK1(up1(x)) -> CHECK1(x)
TOP1(sent1(up1(x))) -> REC1(x)
TOP1(rec1(up1(x))) -> CHECK1(rec1(x))
TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(no1(x)) -> REC1(x)
NO1(up1(x)) -> NO1(x)
SENT1(up1(x)) -> SENT1(x)
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(sent1(x)) -> REC1(x)
CHECK1(sent1(x)) -> CHECK1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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:

TOP1(rec1(up1(x))) -> REC1(x)
CHECK1(no1(x)) -> CHECK1(x)
REC1(bot) -> SENT1(bot)
REC1(rec1(x)) -> SENT1(rec1(x))
TOP1(sent1(up1(x))) -> CHECK1(rec1(x))
CHECK1(rec1(x)) -> CHECK1(x)
REC1(up1(x)) -> REC1(x)
TOP1(no1(up1(x))) -> REC1(x)
TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(no1(x)) -> SENT1(rec1(x))
CHECK1(sent1(x)) -> SENT1(check1(x))
CHECK1(rec1(x)) -> REC1(check1(x))
TOP1(no1(up1(x))) -> CHECK1(rec1(x))
CHECK1(no1(x)) -> NO1(check1(x))
REC1(sent1(x)) -> SENT1(rec1(x))
CHECK1(up1(x)) -> CHECK1(x)
TOP1(sent1(up1(x))) -> REC1(x)
TOP1(rec1(up1(x))) -> CHECK1(rec1(x))
TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(no1(x)) -> REC1(x)
NO1(up1(x)) -> NO1(x)
SENT1(up1(x)) -> SENT1(x)
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
REC1(sent1(x)) -> REC1(x)
CHECK1(sent1(x)) -> CHECK1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 13 less nodes.

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

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

NO1(up1(x)) -> NO1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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.

NO1(up1(x)) -> NO1(x)
Used argument filtering: NO1(x1)  =  x1
up1(x1)  =  up1(x1)
Used ordering: Quasi Precedence: trivial


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

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

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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
          ↳ QDP

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

SENT1(up1(x)) -> SENT1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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.

SENT1(up1(x)) -> SENT1(x)
Used argument filtering: SENT1(x1)  =  x1
up1(x1)  =  up1(x1)
Used ordering: Quasi Precedence: trivial


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

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

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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
          ↳ QDP

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

REC1(up1(x)) -> REC1(x)
REC1(no1(x)) -> REC1(x)
REC1(sent1(x)) -> REC1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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.

REC1(sent1(x)) -> REC1(x)
Used argument filtering: REC1(x1)  =  x1
up1(x1)  =  x1
no1(x1)  =  x1
sent1(x1)  =  sent1(x1)
Used ordering: Quasi Precedence: trivial


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

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

REC1(up1(x)) -> REC1(x)
REC1(no1(x)) -> REC1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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.

REC1(no1(x)) -> REC1(x)
Used argument filtering: REC1(x1)  =  x1
up1(x1)  =  x1
no1(x1)  =  no1(x1)
Used ordering: Quasi Precedence: trivial


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

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

REC1(up1(x)) -> REC1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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.

REC1(up1(x)) -> REC1(x)
Used argument filtering: REC1(x1)  =  x1
up1(x1)  =  up1(x1)
Used ordering: Quasi Precedence: trivial


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

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

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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
            ↳ QDPAfsSolverProof
          ↳ QDP

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

CHECK1(no1(x)) -> CHECK1(x)
CHECK1(up1(x)) -> CHECK1(x)
CHECK1(sent1(x)) -> CHECK1(x)
CHECK1(rec1(x)) -> CHECK1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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(rec1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
no1(x1)  =  x1
up1(x1)  =  x1
sent1(x1)  =  x1
rec1(x1)  =  rec1(x1)
Used ordering: Quasi Precedence: trivial


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

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

CHECK1(no1(x)) -> CHECK1(x)
CHECK1(up1(x)) -> CHECK1(x)
CHECK1(sent1(x)) -> CHECK1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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(sent1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
no1(x1)  =  x1
up1(x1)  =  x1
sent1(x1)  =  sent1(x1)
Used ordering: Quasi Precedence: trivial


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

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

CHECK1(no1(x)) -> CHECK1(x)
CHECK1(up1(x)) -> CHECK1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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(up1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
no1(x1)  =  x1
up1(x1)  =  up1(x1)
Used ordering: Quasi Precedence: trivial


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

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

CHECK1(no1(x)) -> CHECK1(x)

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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(no1(x)) -> CHECK1(x)
Used argument filtering: CHECK1(x1)  =  x1
no1(x1)  =  no1(x1)
Used ordering: Quasi Precedence: trivial


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

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

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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
QDP
            ↳ QDPAfsSolverProof

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

TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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.

TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
Used argument filtering: TOP1(x1)  =  x1
no1(x1)  =  no
check1(x1)  =  x1
rec1(x1)  =  rec
sent1(x1)  =  sent
up1(x1)  =  up
Used ordering: Quasi Precedence: no > [rec, sent, up]


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

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

TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(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.

TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
Used argument filtering: TOP1(x1)  =  x1
sent1(x1)  =  x1
up1(x1)  =  up1(x1)
check1(x1)  =  x1
rec1(x1)  =  rec1(x1)
no1(x1)  =  x1
bot  =  bot
Used ordering: Quasi Precedence: [up_1, rec_1]


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

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

TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))

The TRS R consists of the following rules:

rec1(rec1(x)) -> sent1(rec1(x))
rec1(sent1(x)) -> sent1(rec1(x))
rec1(no1(x)) -> sent1(rec1(x))
rec1(bot) -> up1(sent1(bot))
rec1(up1(x)) -> up1(rec1(x))
sent1(up1(x)) -> up1(sent1(x))
no1(up1(x)) -> up1(no1(x))
top1(rec1(up1(x))) -> top1(check1(rec1(x)))
top1(sent1(up1(x))) -> top1(check1(rec1(x)))
top1(no1(up1(x))) -> top1(check1(rec1(x)))
check1(up1(x)) -> up1(check1(x))
check1(sent1(x)) -> sent1(check1(x))
check1(rec1(x)) -> rec1(check1(x))
check1(no1(x)) -> no1(check1(x))
check1(no1(x)) -> no1(x)

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