Termination w.r.t. Q proof of /tmp/tmpxsGori/LengthOfFiniteLists_nosorts

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:

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

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:

MARK(length(X)) → MARK(X)
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
A__LENGTH(cons(N, L)) → MARK(L)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(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:

MARK(length(X)) → MARK(X)
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
A__LENGTH(cons(N, L)) → MARK(L)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK(length(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

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.

MARK(length(X)) → MARK(X)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
A__AND(tt, X) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

Used ordering: Polynomial interpretation [25,35]:

POL(A__LENGTH(x₁)) = 1 + x_1
POL(a__and(x₁, x₂)) = (2)x_1 + x_2
POL(a__zeros) = 0
POL(mark(x₁)) = x_1
POL(and(x₁, x₂)) = (2)x_1 + x_2
POL(0) = 0
POL(A__AND(x₁, x₂)) = (4)x_1 + (2)x_2
POL(cons(x₁, x₂)) = (2)x_1 + (4)x_2
POL(MARK(x₁)) = (2)x_1
POL(a__length(x₁)) = 1 + x_1
POL(tt) = 1
POL(zeros) = 0
POL(s(x₁)) = x_1
POL(length(x₁)) = 1 + x_1
POL(nil) = 3

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

a__zeros → cons(0, zeros)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(length(X)) → a__length(mark(X))
a__and(tt, X) → mark(X)
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(nil) → nil
mark(tt) → tt
a__zeros → zeros
mark(s(X)) → s(mark(X))
a__length(X) → length(X)
a__and(X1, X2) → and(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

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

A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

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.

MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 4 + (3)x_1
POL(MARK(x₁)) = (4)x_1
POL(s(x₁)) = 4 + (4)x_1
POL(and(x₁, x₂)) = 4 + (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

a__zeros → cons(0, zeros)
a__and(tt, X) → mark(X)
a__length(nil) → 0
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(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.