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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 5 SCCs with 13 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


NO1(up1(x)) -> NO1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(NO1(x1)) = x1   
POL(up1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


SENT1(up1(x)) -> SENT1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SENT1(x1)) = x1   
POL(up1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


REC1(up1(x)) -> REC1(x)
REC1(no1(x)) -> REC1(x)
REC1(sent1(x)) -> REC1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(REC1(x1)) = x1   
POL(no1(x1)) = 1 + x1   
POL(sent1(x1)) = 1 + x1   
POL(up1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


CHECK1(no1(x)) -> CHECK1(x)
CHECK1(up1(x)) -> CHECK1(x)
CHECK1(sent1(x)) -> CHECK1(x)
CHECK1(rec1(x)) -> CHECK1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CHECK1(x1)) = x1   
POL(no1(x1)) = 1 + x1   
POL(rec1(x1)) = 1 + x1   
POL(sent1(x1)) = 1 + x1   
POL(up1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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
            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(no1(up1(x))) -> TOP1(check1(rec1(x)))
The remaining pairs can at least be oriented weakly.

TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = x1   
POL(bot) = 0   
POL(check1(x1)) = x1   
POL(no1(x1)) = 1   
POL(rec1(x1)) = 0   
POL(sent1(x1)) = 0   
POL(up1(x1)) = 0   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(rec1(up1(x))) -> TOP1(check1(rec1(x)))
The remaining pairs can at least be oriented weakly.

TOP1(sent1(up1(x))) -> TOP1(check1(rec1(x)))
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = x1   
POL(bot) = 0   
POL(check1(x1)) = x1   
POL(no1(x1)) = 1 + x1   
POL(rec1(x1)) = 1 + x1   
POL(sent1(x1)) = x1   
POL(up1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
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.