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:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(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:

CHECK(cons(x, y)) → CHECK(y)
TOP(sent(x)) → REST(x)
TOP(sent(x)) → CHECK(rest(x))
TOP(sent(x)) → TOP(check(rest(x)))
CHECK(rest(x)) → REST(check(x))
CHECK(rest(x)) → CHECK(x)
CHECK(cons(x, y)) → CHECK(x)
CHECK(sent(x)) → CHECK(x)

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

CHECK(cons(x, y)) → CHECK(y)
TOP(sent(x)) → REST(x)
TOP(sent(x)) → CHECK(rest(x))
TOP(sent(x)) → TOP(check(rest(x)))
CHECK(rest(x)) → REST(check(x))
CHECK(rest(x)) → CHECK(x)
CHECK(cons(x, y)) → CHECK(x)
CHECK(sent(x)) → CHECK(x)

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

TOP(sent(x)) → REST(x)
CHECK(cons(x, y)) → CHECK(y)
TOP(sent(x)) → CHECK(rest(x))
TOP(sent(x)) → TOP(check(rest(x)))
CHECK(rest(x)) → REST(check(x))
CHECK(rest(x)) → CHECK(x)
CHECK(sent(x)) → CHECK(x)
CHECK(cons(x, y)) → CHECK(x)

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

CHECK(cons(x, y)) → CHECK(y)
CHECK(rest(x)) → CHECK(x)
CHECK(cons(x, y)) → CHECK(x)
CHECK(sent(x)) → CHECK(x)

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(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.


CHECK(cons(x, y)) → CHECK(y)
CHECK(cons(x, y)) → CHECK(x)
The remaining pairs can at least be oriented weakly.

CHECK(rest(x)) → CHECK(x)
CHECK(sent(x)) → CHECK(x)
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
rest(x1)  =  x1
sent(x1)  =  x1

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

CHECK(rest(x)) → CHECK(x)
CHECK(sent(x)) → CHECK(x)

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(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.


CHECK(sent(x)) → CHECK(x)
The remaining pairs can at least be oriented weakly.

CHECK(rest(x)) → CHECK(x)
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
rest(x1)  =  x1
sent(x1)  =  sent(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

CHECK(rest(x)) → CHECK(x)

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(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.


CHECK(rest(x)) → CHECK(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
CHECK(x1)  =  x1
rest(x1)  =  rest(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ PisEmptyProof
              ↳ QDP

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

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

TOP(sent(x)) → TOP(check(rest(x)))

The TRS R consists of the following rules:

top(sent(x)) → top(check(rest(x)))
rest(nil) → sent(nil)
rest(cons(x, y)) → sent(y)
check(sent(x)) → sent(check(x))
check(rest(x)) → rest(check(x))
check(cons(x, y)) → cons(check(x), y)
check(cons(x, y)) → cons(x, check(y))
check(cons(x, y)) → cons(x, y)

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