Termination w.r.t. Q proof of /tmp/tmpoYIFH3/OvConsOS_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)))
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

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)))
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

The following Q TRS is given: 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)))
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

a__and(tt, X) → mark(X)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__and(x₁, x₂)) = x₁ + x₂
POL(a__length(x₁)) = x₁
POL(a__take(x₁, x₂)) = x₁ + x₂
POL(a__zeros) = 0
POL(and(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = 2·x₁
POL(take(x₁, x₂)) = x₁ + x₂
POL(tt) = 1
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

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

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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

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

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

a__take(0, IL) → nil

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__and(x₁, x₂)) = 2·x₁ + 2·x₂
POL(a__length(x₁)) = x₁
POL(a__take(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(a__zeros) = 0
POL(and(x₁, x₂)) = 2·x₁ + 2·x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

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

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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

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

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

a__length(nil) → 0

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__and(x₁, x₂)) = 2·x₁ + x₂
POL(a__length(x₁)) = x₁
POL(a__take(x₁, x₂)) = x₁ + 2·x₂
POL(a__zeros) = 0
POL(and(x₁, x₂)) = 2·x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 2
POL(s(x₁)) = 2·x₁
POL(take(x₁, x₂)) = x₁ + 2·x₂
POL(tt) = 2
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
MARK(length(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
A__TAKE(s(M), cons(N, IL)) → MARK(N)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

MARK(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
A__LENGTH(cons(N, L)) → MARK(L)
MARK(length(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
A__TAKE(s(M), cons(N, IL)) → MARK(N)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

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

MARK(take(X1, X2)) → MARK(X2)
MARK(length(X)) → MARK(X)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(take(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__TAKE(s(M), cons(N, IL)) → MARK(N)
A__LENGTH(cons(N, L)) → MARK(L)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(and(X1, X2)) → MARK(X1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__LENGTH(x₁)) = x₁
POL(A__TAKE(x₁, x₂)) = x₁ + x₂
POL(MARK(x₁)) = x₁
POL(a__and(x₁, x₂)) = 2 + x₁ + x₂
POL(a__length(x₁)) = 2·x₁
POL(a__take(x₁, x₂)) = x₁ + x₂
POL(a__zeros) = 0
POL(and(x₁, x₂)) = 2 + x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = 2·x₁
POL(take(x₁, x₂)) = x₁ + x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

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

MARK(length(X)) → MARK(X)
MARK(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(take(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__TAKE(s(M), cons(N, IL)) → MARK(N)
A__LENGTH(cons(N, L)) → MARK(L)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__LENGTH(x₁)) = 1 + 2·x₁
POL(A__TAKE(x₁, x₂)) = 2·x₁ + x₂
POL(MARK(x₁)) = 2·x₁
POL(a__and(x₁, x₂)) = x₁ + x₂
POL(a__length(x₁)) = 1 + x₁
POL(a__take(x₁, x₂)) = x₁ + 2·x₂
POL(a__zeros) = 0
POL(and(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = x₁ + 2·x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

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

MARK(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
A__TAKE(s(M), cons(N, IL)) → MARK(N)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                  ↳ 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__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__LENGTH(cons(N, L)) → A__LENGTH(mark(L)) at position [0] we obtained the following new rules:

A__LENGTH(cons(y0, s(x0))) → A__LENGTH(s(mark(x0)))
A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))
A__LENGTH(cons(y0, tt)) → A__LENGTH(tt)
A__LENGTH(cons(y0, 0)) → A__LENGTH(0)
A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, nil)) → A__LENGTH(nil)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                  ↳ QDP

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

A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))
A__LENGTH(cons(y0, s(x0))) → A__LENGTH(s(mark(x0)))
A__LENGTH(cons(y0, tt)) → A__LENGTH(tt)
A__LENGTH(cons(y0, 0)) → A__LENGTH(0)
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, nil)) → A__LENGTH(nil)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                  ↳ QDP

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

A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros)
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__LENGTH(cons(y0, zeros)) → A__LENGTH(a__zeros) at position [0] we obtained the following new rules:

A__LENGTH(cons(y0, zeros)) → A__LENGTH(zeros)
A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                  ↳ QDP

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

A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, zeros)) → A__LENGTH(zeros)
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ AND

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                    ↳ QDP

                                  ↳ QDP

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ AND

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                    ↳ QDP

                                  ↳ QDP

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

R is empty.
The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ AND

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Instantiation

                                                    ↳ QDP

                                  ↳ QDP

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

A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule A__LENGTH(cons(y0, zeros)) → A__LENGTH(cons(0, zeros)) we obtained the following new rules:

A__LENGTH(cons(0, zeros)) → A__LENGTH(cons(0, zeros))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ AND

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Instantiation

                                                                ↳ QDP

                                                                  ↳ NonTerminationProof

                                                    ↳ QDP

                                  ↳ QDP

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

A__LENGTH(cons(0, zeros)) → A__LENGTH(cons(0, zeros))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

A__LENGTH(cons(0, zeros)) → A__LENGTH(cons(0, zeros))

The TRS R consists of the following rules:none

s = A__LENGTH(cons(0, zeros)) evaluates to t =A__LENGTH(cons(0, zeros))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from A__LENGTH(cons(0, zeros)) to A__LENGTH(cons(0, zeros)).

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ AND

                                                    ↳ QDP

                                                    ↳ QDP

                                                      ↳ RuleRemovalProof

                                  ↳ QDP

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

A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))
A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

A__LENGTH(cons(y0, and(x0, x1))) → A__LENGTH(a__and(mark(x0), x1))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__LENGTH(x₁)) = x₁
POL(a__and(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(a__length(x₁)) = x₁
POL(a__take(x₁, x₂)) = x₁ + 2·x₂
POL(a__zeros) = 0
POL(and(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = 2·x₁
POL(take(x₁, x₂)) = x₁ + 2·x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ AND

                                                    ↳ QDP

                                                    ↳ QDP

                                                      ↳ RuleRemovalProof

                                                        ↳ QDP

                                                          ↳ QDPOrderProof

                                  ↳ QDP

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

A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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.

A__LENGTH(cons(y0, length(x0))) → A__LENGTH(a__length(mark(x0)))

The remaining pairs can at least be oriented weakly.

A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0
POL(A__LENGTH(x₁)) = x₁
POL(a__and(x₁, x₂)) = x₂
POL(a__length(x₁)) = 0
POL(a__take(x₁, x₂)) = 1
POL(a__zeros) = 0
POL(and(x₁, x₂)) = x₂
POL(cons(x₁, x₂)) = 1
POL(length(x₁)) = 0
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = 0
POL(take(x₁, x₂)) = 1
POL(tt) = 0
POL(zeros) = 0

The following usable rules [17] were oriented:

a__take(X1, X2) → take(X1, X2)
a__length(X) → length(X)
mark(length(X)) → a__length(mark(X))
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → s(a__length(mark(L)))
a__and(X1, X2) → and(X1, X2)
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ AND

                                                    ↳ QDP

                                                    ↳ QDP

                                                      ↳ RuleRemovalProof

                                                        ↳ QDP

                                                          ↳ QDPOrderProof

                                                            ↳ QDP

                                                              ↳ QDPOrderProof

                                  ↳ QDP

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

A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))
A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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.

A__LENGTH(cons(y0, cons(x0, x1))) → A__LENGTH(cons(mark(x0), x1))

The remaining pairs can at least be oriented weakly.

A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0
POL(A__LENGTH(x₁)) = x₁
POL(a__and(x₁, x₂)) = 1
POL(a__length(x₁)) = 0
POL(a__take(x₁, x₂)) = 1
POL(a__zeros) = 0
POL(and(x₁, x₂)) = 1
POL(cons(x₁, x₂)) = 1 + x₂
POL(length(x₁)) = 0
POL(mark(x₁)) = 1 + x₁
POL(nil) = 0
POL(s(x₁)) = 0
POL(take(x₁, x₂)) = 0
POL(tt) = 0
POL(zeros) = 0

The following usable rules [17] were oriented:

a__take(X1, X2) → take(X1, X2)
a__length(X) → length(X)
mark(length(X)) → a__length(mark(X))
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → s(a__length(mark(L)))
a__and(X1, X2) → and(X1, X2)
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ AND

                                                    ↳ QDP

                                                    ↳ QDP

                                                      ↳ RuleRemovalProof

                                                        ↳ QDP

                                                          ↳ QDPOrderProof

                                                            ↳ QDP

                                                              ↳ QDPOrderProof

                                                                ↳ QDP

                                  ↳ QDP

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

A__LENGTH(cons(y0, take(x0, x1))) → A__LENGTH(a__take(mark(x0), mark(x1)))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

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

MARK(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__TAKE(s(M), cons(N, IL)) → MARK(N)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
a__take(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(zeros) → a__zeros
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
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)
a__take(X1, X2) → take(X1, X2)

The set Q consists of the following terms:

mark(0)
mark(tt)
a__length(x0)
mark(zeros)
mark(cons(x0, x1))
a__and(x0, x1)
a__take(x0, x1)
a__zeros
mark(take(x0, x1))
mark(and(x0, x1))
mark(nil)
mark(s(x0))
mark(length(x0))

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