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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs with 3 less nodes.

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


The following pairs can be oriented strictly and are deleted.


CHECK1(cons2(x, y)) -> CHECK1(y)
CHECK1(rest1(x)) -> CHECK1(x)
CHECK1(cons2(x, y)) -> CHECK1(x)
CHECK1(sent1(x)) -> CHECK1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CHECK1(x1)) = 3·x1 + 3·x12   
POL(cons2(x1, x2)) = 3 + 2·x1 + x2   
POL(rest1(x1)) = 3 + x1   
POL(sent1(x1)) = 3 + 3·x1 + 3·x12   

The following usable rules [14] were oriented: none



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