Termination proof of /tmp/tmpr5lVSF/OvConsOS

Termination of the following Term Rewriting System could not be shown:

Context-sensitive rewrite system:
The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The replacement map contains the following entries:

zeros: empty set
cons: {1}
0: empty set
and: {1}
tt: empty set
length: {1}
nil: empty set
s: {1}
take: {1, 2}

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

Context-sensitive rewrite system:
The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The replacement map contains the following entries:

zeros: empty set
cons: {1}
0: empty set
and: {1}
tt: empty set
length: {1}
nil: empty set
s: {1}
take: {1, 2}

The CSR is orthogonal. By [10] we can switch to innermost.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

Context-sensitive rewrite system:
The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The replacement map contains the following entries:

zeros: empty set
cons: {1}
0: empty set
and: {1}
tt: empty set
length: {1}
nil: empty set
s: {1}
take: {1, 2}

Innermost Strategy.

Using Improved CS-DPs we result in the following initial Q-CSDP problem.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

Q-restricted context-sensitive dependency pair problem:
The symbols in {length, s, take, LENGTH, TAKE} are replacing on all positions.
For all symbols f in {cons, and, AND} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The ordinary context-sensitive dependency pairs DP_o are:

LENGTH(cons(N, L)) → LENGTH(L)

The collapsing dependency pairs are DP_c:

AND(tt, X) → X
LENGTH(cons(N, L)) → L

The hidden terms of R are:

zeros
take(M, IL)

Every hiding context is built from:

take on positions {1, 2}

Hence, the new unhiding pairs DP_u are :

AND(tt, X) → U(X)
LENGTH(cons(N, L)) → U(L)
U(take(x_0, x_1)) → U(x_0)
U(take(x_0, x_1)) → U(x_1)
U(zeros) → ZEROS
U(take(M, IL)) → TAKE(M, IL)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
and(tt, x0)
length(nil)
length(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))

The approximation of the Context-Sensitive Dependency Graph contains 2 SCCs with 4 less nodes.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ AND

              ↳ QCSDP

                ↳ QCSDPSubtermProof

              ↳ QCSDP

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

Q-restricted context-sensitive dependency pair problem:
The symbols in {length, s, take} are replacing on all positions.
For all symbols f in {cons, and} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The TRS P consists of the following rules:

U(take(x_0, x_1)) → U(x_0)
U(take(x_0, x_1)) → U(x_1)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
and(tt, x0)
length(nil)
length(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))

We use the subterm processor [20].

The following pairs can be oriented strictly and are deleted.

U(take(x_0, x_1)) → U(x_0)
U(take(x_0, x_1)) → U(x_1)

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

Subterm Order

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ AND

              ↳ QCSDP

                ↳ QCSDPSubtermProof

                  ↳ QCSDP

                    ↳ PIsEmptyProof

              ↳ QCSDP

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

Q-restricted context-sensitive dependency pair problem:
The symbols in {length, s, take} are replacing on all positions.
For all symbols f in {cons, and} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
and(tt, x0)
length(nil)
length(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))

The TRS P is empty. Hence, there is no (P,Q,R,µ)-chain.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ AND

              ↳ QCSDP

              ↳ QCSDP

                ↳ ConvertedToQDPProblemProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

Q-restricted context-sensitive dependency pair problem:
The symbols in {length, s, take, LENGTH} are replacing on all positions.
For all symbols f in {cons, and} we have µ(f) = {1}.

The TRS P consists of the following rules:

LENGTH(cons(N, L)) → LENGTH(L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
and(tt, x0)
length(nil)
length(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))

Converted QDP Problem, but could not keep Q or minimality.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ AND

              ↳ QCSDP

              ↳ QCSDP

                ↳ ConvertedToQDPProblemProof

                  ↳ QDP

                    ↳ NonTerminationProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

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

LENGTH(cons(N, L)) → LENGTH(L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

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

The TRS P consists of the following rules:

LENGTH(cons(N, L)) → LENGTH(L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

s = LENGTH(zeros) evaluates to t =LENGTH(zeros)

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

Matcher: [ ]
Semiunifier: [ ]

Rewriting sequence

LENGTH(zeros) → LENGTH(cons(0, zeros))
with rule zeros → cons(0, zeros) at position [0] and matcher [ ]

LENGTH(cons(0, zeros)) → LENGTH(zeros)
with rule LENGTH(cons(N, L)) → LENGTH(L)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.

We applied the Incomplete Giesl Middeldorp transformation [11] to transform the context-sensitive TRS to a usual TRS.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

  ↳ Trivial-Transformation

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

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
andActive(tt, X) → mark(X)
lengthActive(nil) → 0
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(0, IL) → nil
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

Q is empty.

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

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
andActive(tt, X) → mark(X)
lengthActive(nil) → 0
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(0, IL) → nil
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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

andActive(tt, X) → mark(X)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(and(x₁, x₂)) = 2·x₁ + x₂
POL(andActive(x₁, x₂)) = 2·x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = 2·x₁
POL(lengthActive(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = x₁ + x₂
POL(takeActive(x₁, x₂)) = x₁ + x₂
POL(tt) = 2
POL(zeros) = 0
POL(zerosActive) = 0

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

  ↳ Trivial-Transformation

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

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(nil) → 0
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(0, IL) → nil
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

Q is empty.

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

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(nil) → 0
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(0, IL) → nil
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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

takeActive(0, IL) → nil

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(and(x₁, x₂)) = x₁ + 2·x₂
POL(andActive(x₁, x₂)) = x₁ + 2·x₂
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = x₁
POL(lengthActive(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(takeActive(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(tt) = 0
POL(zeros) = 0
POL(zerosActive) = 0

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

  ↳ Trivial-Transformation

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

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(nil) → 0
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

Q is empty.

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

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(nil) → 0
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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

lengthActive(nil) → 0

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(and(x₁, x₂)) = 2·x₁ + x₂
POL(andActive(x₁, x₂)) = 2·x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = 2·x₁
POL(lengthActive(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 1
POL(s(x₁)) = 2·x₁
POL(take(x₁, x₂)) = x₁ + x₂
POL(takeActive(x₁, x₂)) = x₁ + x₂
POL(tt) = 0
POL(zeros) = 0
POL(zerosActive) = 0

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

  ↳ Trivial-Transformation

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

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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:

LENGTHACTIVE(cons(N, L)) → MARK(L)
MARK(length(x1)) → LENGTHACTIVE(mark(x1))
MARK(take(x1, x2)) → MARK(x2)
LENGTHACTIVE(cons(N, L)) → LENGTHACTIVE(mark(L))
TAKEACTIVE(s(M), cons(N, IL)) → MARK(N)
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
MARK(zeros) → ZEROSACTIVE
MARK(and(x1, x2)) → ANDACTIVE(mark(x1), x2)
MARK(length(x1)) → MARK(x1)
MARK(and(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(N, L)) → MARK(L)
MARK(length(x1)) → LENGTHACTIVE(mark(x1))
MARK(take(x1, x2)) → MARK(x2)
LENGTHACTIVE(cons(N, L)) → LENGTHACTIVE(mark(L))
TAKEACTIVE(s(M), cons(N, IL)) → MARK(N)
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))
MARK(zeros) → ZEROSACTIVE
MARK(and(x1, x2)) → ANDACTIVE(mark(x1), x2)
MARK(length(x1)) → MARK(x1)
MARK(and(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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 2 less nodes.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(N, L)) → MARK(L)
MARK(take(x1, x2)) → MARK(x2)
MARK(length(x1)) → MARK(x1)
MARK(length(x1)) → LENGTHACTIVE(mark(x1))
TAKEACTIVE(s(M), cons(N, IL)) → MARK(N)
LENGTHACTIVE(cons(N, L)) → LENGTHACTIVE(mark(L))
MARK(s(x1)) → MARK(x1)
MARK(and(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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(take(x1, x2)) → MARK(x2)
TAKEACTIVE(s(M), cons(N, IL)) → MARK(N)
MARK(and(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → MARK(x1)

The remaining pairs can at least be oriented weakly.

LENGTHACTIVE(cons(N, L)) → MARK(L)
MARK(length(x1)) → MARK(x1)
MARK(length(x1)) → LENGTHACTIVE(mark(x1))
LENGTHACTIVE(cons(N, L)) → LENGTHACTIVE(mark(L))
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(LENGTHACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(TAKEACTIVE(x₁, x₂)) = 1 + x₂
POL(and(x₁, x₂)) = 1 + x₁
POL(andActive(x₁, x₂)) = 1 + x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(length(x₁)) = x₁
POL(lengthActive(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 1 + x₁ + x₂
POL(takeActive(x₁, x₂)) = 1 + x₁ + x₂
POL(tt) = 0
POL(zeros) = 0
POL(zerosActive) = 0

The following usable rules [17] were oriented:

takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
zerosActive → cons(0, zeros)
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
mark(0) → 0
mark(tt) → tt
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
mark(zeros) → zerosActive

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(N, L)) → MARK(L)
MARK(length(x1)) → LENGTHACTIVE(mark(x1))
MARK(length(x1)) → MARK(x1)
LENGTHACTIVE(cons(N, L)) → LENGTHACTIVE(mark(L))
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(take(x1, x2)) → TAKEACTIVE(mark(x1), mark(x2))

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(N, L)) → MARK(L)
MARK(length(x1)) → MARK(x1)
MARK(length(x1)) → LENGTHACTIVE(mark(x1))
LENGTHACTIVE(cons(N, L)) → LENGTHACTIVE(mark(L))
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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(x1)) → MARK(x1)
MARK(length(x1)) → LENGTHACTIVE(mark(x1))

The remaining pairs can at least be oriented weakly.

LENGTHACTIVE(cons(N, L)) → MARK(L)
LENGTHACTIVE(cons(N, L)) → LENGTHACTIVE(mark(L))
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(LENGTHACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(and(x₁, x₂)) = 1 + x₂
POL(andActive(x₁, x₂)) = 1 + x₂
POL(cons(x₁, x₂)) = x₁ + x₂
POL(length(x₁)) = 1 + x₁
POL(lengthActive(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = x₂
POL(takeActive(x₁, x₂)) = x₂
POL(tt) = 0
POL(zeros) = 0
POL(zerosActive) = 0

The following usable rules [17] were oriented:

takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
zerosActive → cons(0, zeros)
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
mark(0) → 0
mark(tt) → tt
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
mark(zeros) → zerosActive

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(N, L)) → MARK(L)
LENGTHACTIVE(cons(N, L)) → LENGTHACTIVE(mark(L))
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                          ↳ QDP

  ↳ Trivial-Transformation

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

MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QDPSizeChangeProof

                                          ↳ QDP

  ↳ Trivial-Transformation

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

MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

MARK(s(x1)) → MARK(x1)
The graph contains the following edges 1 > 1
MARK(cons(x1, x2)) → MARK(x1)
The graph contains the following edges 1 > 1

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

  ↳ Trivial-Transformation

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

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

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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

LENGTHACTIVE(cons(y0, zeros)) → LENGTHACTIVE(zerosActive)
LENGTHACTIVE(cons(y0, cons(x0, x1))) → LENGTHACTIVE(cons(mark(x0), x1))
LENGTHACTIVE(cons(y0, s(x0))) → LENGTHACTIVE(s(mark(x0)))
LENGTHACTIVE(cons(y0, length(x0))) → LENGTHACTIVE(lengthActive(mark(x0)))
LENGTHACTIVE(cons(y0, and(x0, x1))) → LENGTHACTIVE(andActive(mark(x0), x1))
LENGTHACTIVE(cons(y0, nil)) → LENGTHACTIVE(nil)
LENGTHACTIVE(cons(y0, 0)) → LENGTHACTIVE(0)
LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))
LENGTHACTIVE(cons(y0, tt)) → LENGTHACTIVE(tt)

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(y0, zeros)) → LENGTHACTIVE(zerosActive)
LENGTHACTIVE(cons(y0, cons(x0, x1))) → LENGTHACTIVE(cons(mark(x0), x1))
LENGTHACTIVE(cons(y0, s(x0))) → LENGTHACTIVE(s(mark(x0)))
LENGTHACTIVE(cons(y0, length(x0))) → LENGTHACTIVE(lengthActive(mark(x0)))
LENGTHACTIVE(cons(y0, and(x0, x1))) → LENGTHACTIVE(andActive(mark(x0), x1))
LENGTHACTIVE(cons(y0, nil)) → LENGTHACTIVE(nil)
LENGTHACTIVE(cons(y0, 0)) → LENGTHACTIVE(0)
LENGTHACTIVE(cons(y0, tt)) → LENGTHACTIVE(tt)
LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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 4 less nodes.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ Narrowing

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(y0, zeros)) → LENGTHACTIVE(zerosActive)
LENGTHACTIVE(cons(y0, cons(x0, x1))) → LENGTHACTIVE(cons(mark(x0), x1))
LENGTHACTIVE(cons(y0, length(x0))) → LENGTHACTIVE(lengthActive(mark(x0)))
LENGTHACTIVE(cons(y0, and(x0, x1))) → LENGTHACTIVE(andActive(mark(x0), x1))
LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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

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

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(y0, cons(x0, x1))) → LENGTHACTIVE(cons(mark(x0), x1))
LENGTHACTIVE(cons(y0, length(x0))) → LENGTHACTIVE(lengthActive(mark(x0)))
LENGTHACTIVE(cons(y0, zeros)) → LENGTHACTIVE(cons(0, zeros))
LENGTHACTIVE(cons(y0, and(x0, x1))) → LENGTHACTIVE(andActive(mark(x0), x1))
LENGTHACTIVE(cons(y0, zeros)) → LENGTHACTIVE(zeros)
LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                                            ↳ QDP

  ↳ Trivial-Transformation

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

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

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                              ↳ UsableRulesProof

                                                                ↳ QDP

                                                            ↳ QDP

  ↳ Trivial-Transformation

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

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

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

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ QDPOrderProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(y0, cons(x0, x1))) → LENGTHACTIVE(cons(mark(x0), x1))
LENGTHACTIVE(cons(y0, length(x0))) → LENGTHACTIVE(lengthActive(mark(x0)))
LENGTHACTIVE(cons(y0, and(x0, x1))) → LENGTHACTIVE(andActive(mark(x0), x1))
LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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.

LENGTHACTIVE(cons(y0, length(x0))) → LENGTHACTIVE(lengthActive(mark(x0)))
LENGTHACTIVE(cons(y0, and(x0, x1))) → LENGTHACTIVE(andActive(mark(x0), x1))

The remaining pairs can at least be oriented weakly.

LENGTHACTIVE(cons(y0, cons(x0, x1))) → LENGTHACTIVE(cons(mark(x0), x1))
LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(LENGTHACTIVE(x₁)) = x₁
POL(and(x₁, x₂)) = 0
POL(andActive(x₁, x₂)) = 0
POL(cons(x₁, x₂)) = 1
POL(length(x₁)) = 0
POL(lengthActive(x₁)) = 0
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = 0
POL(take(x₁, x₂)) = 1
POL(takeActive(x₁, x₂)) = 1
POL(tt) = 0
POL(zeros) = 0
POL(zerosActive) = 0

The following usable rules [17] were oriented:

takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(s(x1)) → s(mark(x1))
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(and(x1, x2)) → andActive(mark(x1), x2)

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ QDPOrderProof

                                                                ↳ QDP

                                                                  ↳ QDPOrderProof

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(y0, cons(x0, x1))) → LENGTHACTIVE(cons(mark(x0), x1))
LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

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.

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

The remaining pairs can at least be oriented weakly.

LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(LENGTHACTIVE(x₁)) = x₁
POL(and(x₁, x₂)) = 1
POL(andActive(x₁, x₂)) = 1
POL(cons(x₁, x₂)) = 1 + x₂
POL(length(x₁)) = 0
POL(lengthActive(x₁)) = 0
POL(mark(x₁)) = 1 + x₁
POL(nil) = 1
POL(s(x₁)) = 0
POL(take(x₁, x₂)) = x₂
POL(takeActive(x₁, x₂)) = x₂
POL(tt) = 1
POL(zeros) = 0
POL(zerosActive) = 1

The following usable rules [17] were oriented:

takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))
mark(s(x1)) → s(mark(x1))
mark(nil) → nil
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
zerosActive → cons(0, zeros)
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
mark(0) → 0
mark(tt) → tt
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
mark(zeros) → zerosActive

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ AND

                                          ↳ QDP

                                          ↳ QDP

                                            ↳ Narrowing

                                              ↳ QDP

                                                ↳ DependencyGraphProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ AND

                                                            ↳ QDP

                                                            ↳ QDP

                                                              ↳ QDPOrderProof

                                                                ↳ QDP

                                                                  ↳ QDPOrderProof

                                                                    ↳ QDP

  ↳ Trivial-Transformation

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

LENGTHACTIVE(cons(y0, take(x0, x1))) → LENGTHACTIVE(takeActive(mark(x0), mark(x1)))

The TRS R consists of the following rules:

mark(zeros) → zerosActive
zerosActive → zeros
mark(and(x1, x2)) → andActive(mark(x1), x2)
andActive(x1, x2) → and(x1, x2)
mark(length(x1)) → lengthActive(mark(x1))
lengthActive(x1) → length(x1)
mark(take(x1, x2)) → takeActive(mark(x1), mark(x2))
takeActive(x1, x2) → take(x1, x2)
mark(cons(x1, x2)) → cons(mark(x1), x2)
mark(0) → 0
mark(tt) → tt
mark(nil) → nil
mark(s(x1)) → s(mark(x1))
zerosActive → cons(0, zeros)
lengthActive(cons(N, L)) → s(lengthActive(mark(L)))
takeActive(s(M), cons(N, IL)) → cons(mark(N), take(M, IL))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We applied the Trivial transformation to transform the context-sensitive TRS to a usual TRS.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

Q is empty.

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

zeros → cons(0, zeros)
and(tt, X) → X
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

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

and(tt, X) → X

Used ordering:
Polynomial interpretation [25]:

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

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

zeros → cons(0, zeros)
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

Q is empty.

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

zeros → cons(0, zeros)
length(nil) → 0
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

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

length(nil) → 0

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = 2·x₁
POL(nil) = 2
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(zeros) = 0

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

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

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

Q is empty.

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

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

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

take(0, IL) → nil

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = 2·x₁
POL(nil) = 1
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(zeros) = 0

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

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

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

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

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LENGTH(cons(N, L)) → LENGTH(L)
ZEROS → ZEROS
TAKE(s(M), cons(N, IL)) → TAKE(M, IL)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

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

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

LENGTH(cons(N, L)) → LENGTH(L)
ZEROS → ZEROS
TAKE(s(M), cons(N, IL)) → TAKE(M, IL)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                              ↳ QDP

                              ↳ QDP

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

TAKE(s(M), cons(N, IL)) → TAKE(M, IL)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

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.

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                              ↳ QDP

                              ↳ QDP

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

TAKE(s(M), cons(N, IL)) → TAKE(M, IL)

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

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

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.

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ QDPSizeChangeProof

                              ↳ QDP

                              ↳ QDP

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

TAKE(s(M), cons(N, IL)) → TAKE(M, IL)

From the DPs we obtained the following set of size-change graphs:

TAKE(s(M), cons(N, IL)) → TAKE(M, IL)
The graph contains the following edges 1 > 1, 2 > 2

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                              ↳ QDP

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

LENGTH(cons(N, L)) → LENGTH(L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                              ↳ QDP

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

LENGTH(cons(N, L)) → LENGTH(L)

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

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

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.

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ QDPSizeChangeProof

                              ↳ QDP

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

LENGTH(cons(N, L)) → LENGTH(L)

From the DPs we obtained the following set of size-change graphs:

LENGTH(cons(N, L)) → LENGTH(L)
The graph contains the following edges 1 > 1

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

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

ZEROS → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, zeros)
length(cons(N, L)) → s(length(L))
take(s(M), cons(N, IL)) → cons(N, take(M, IL))

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

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

ZEROS → ZEROS

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

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

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.

zeros
length(cons(x0, x1))
take(s(x0), cons(x1, x2))

↳ CSR

  ↳ CSRInnermostProof

  ↳ Incomplete Giesl Middeldorp-Transformation

  ↳ Trivial-Transformation

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ Overlay + Local Confluence

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

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

ZEROS → ZEROS

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