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 [1,15] we result in the following initial DP problem:
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(sent(x)) → CHECK(x)
CHECK(rest(x)) → CHECK(x)
CHECK(cons(x, y)) → CHECK(x)
CHECK(rest(x)) → REST(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
      ↳ 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(sent(x)) → CHECK(x)
CHECK(rest(x)) → CHECK(x)
CHECK(cons(x, y)) → CHECK(x)
CHECK(rest(x)) → REST(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 [15,17,22] 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:

CHECK(cons(x, y)) → CHECK(y)
CHECK(sent(x)) → CHECK(x)
CHECK(rest(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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


CHECK(cons(x, y)) → CHECK(y)
CHECK(sent(x)) → CHECK(x)
CHECK(rest(x)) → CHECK(x)
CHECK(cons(x, y)) → CHECK(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 2 + (4)x_1 + x_2   
POL(CHECK(x1)) = (4)x_1   
POL(sent(x1)) = 3 + x_1   
POL(rest(x1)) = 4 + (2)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] 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:

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
      ↳ 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.