Termination proof of /tmp/tmpSFlgzw/LengthOfFiniteLists

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)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(nil) → 0
length(cons(N, L)) → U11(tt, L)

The replacement map contains the following entries:

zeros: empty set
cons: {1}
0: empty set
U11: {1}
tt: empty set
U12: {1}
s: {1}
length: {1}
nil: empty set

↳ CSR

  ↳ CSRInnermostProof

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

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(nil) → 0
length(cons(N, L)) → U11(tt, L)

The replacement map contains the following entries:

zeros: empty set
cons: {1}
0: empty set
U11: {1}
tt: empty set
U12: {1}
s: {1}
length: {1}
nil: empty set

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(nil) → 0
length(cons(N, L)) → U11(tt, L)

The replacement map contains the following entries:

zeros: empty set
cons: {1}
0: empty set
U11: {1}
tt: empty set
U12: {1}
s: {1}
length: {1}
nil: empty set

Innermost Strategy.

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

Q-restricted context-sensitive dependency pair problem:
The symbols in {s, length, LENGTH} are replacing on all positions.
For all symbols f in {cons, U11, U12, U12¹, U11¹} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The ordinary context-sensitive dependency pairs DP_o are:

U11¹(tt, L) → U12¹(tt, L)
U12¹(tt, L) → LENGTH(L)
LENGTH(cons(N, L)) → U11¹(tt, L)

The collapsing dependency pairs are DP_c:

U12¹(tt, L) → L

The hidden terms of R are:

zeros

Every hiding context is built from:none

Hence, the new unhiding pairs DP_u are :

U12¹(tt, L) → U(L)
U(zeros) → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(nil) → 0
length(cons(N, L)) → U11(tt, L)

The set Q consists of the following terms:

zeros
U11(tt, x0)
U12(tt, x0)
length(nil)
length(cons(x0, x1))

The approximation of the Context-Sensitive Dependency Graph contains 1 SCC with 2 less nodes.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ QCSDP

              ↳ ConvertedToQDPProblemProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

U12¹(tt, L) → LENGTH(L)
LENGTH(cons(N, L)) → U11¹(tt, L)
U11¹(tt, L) → U12¹(tt, L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(nil) → 0
length(cons(N, L)) → U11(tt, L)

The set Q consists of the following terms:

zeros
U11(tt, x0)
U12(tt, x0)
length(nil)
length(cons(x0, x1))

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ QCSDP

              ↳ ConvertedToQDPProblemProof

                ↳ QDP

                  ↳ RuleRemovalProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(cons(N, L)) → U11¹(tt, L)
U11¹(tt, L) → U12¹(tt, L)
U12¹(tt, L) → LENGTH(L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(nil) → 0
length(cons(N, L)) → U11(tt, L)

Q is empty.
We have to consider all (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 rules of the TRS R:

length(nil) → 0

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(LENGTH(x₁)) = x₁
POL(U11(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U11¹(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U12(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U12¹(x₁, x₂)) = x₁ + 2·x₂
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = x₁
POL(nil) = 1
POL(s(x₁)) = 2·x₁
POL(tt) = 0
POL(zeros) = 0

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

        ↳ QCSDP

          ↳ QCSDependencyGraphProof

            ↳ QCSDP

              ↳ ConvertedToQDPProblemProof

                ↳ QDP

                  ↳ RuleRemovalProof

                    ↳ QDP

                      ↳ NonTerminationProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(cons(N, L)) → U11¹(tt, L)
U12¹(tt, L) → LENGTH(L)
U11¹(tt, L) → U12¹(tt, L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(cons(N, L)) → U11(tt, L)

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)) → U11¹(tt, L)
U12¹(tt, L) → LENGTH(L)
U11¹(tt, L) → U12¹(tt, L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(cons(N, L)) → U11(tt, L)

s = U11¹(tt, zeros) evaluates to t =U11¹(tt, zeros)

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

Matcher: [ ]
Semiunifier: [ ]

Rewriting sequence

U11¹(tt, zeros) → U11¹(tt, cons(0, zeros))
with rule zeros → cons(0, zeros) at position [1] and matcher [ ]

U11¹(tt, cons(0, zeros)) → U12¹(tt, cons(0, zeros))
with rule U11¹(tt, L') → U12¹(tt, L') at position [] and matcher [L' / cons(0, zeros)]

U12¹(tt, cons(0, zeros)) → LENGTH(cons(0, zeros))
with rule U12¹(tt, L') → LENGTH(L') at position [] and matcher [L' / cons(0, zeros)]

LENGTH(cons(0, zeros)) → U11¹(tt, zeros)
with rule LENGTH(cons(N, L)) → U11¹(tt, 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 Zantema transformation [34] to transform the context-sensitive TRS to a usual TRS.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

Q is empty.

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

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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(U11(x₁, x₂)) = x₁ + 2·x₂
POL(U12(x₁, x₂)) = x₁ + 2·x₂
POL(a(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = x₁
POL(nil) = 2
POL(s(x₁)) = x₁
POL(tt) = 0
POL(zeros) = 0
POL(zerosInact) = 0

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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:

A(zerosInact) → ZEROS
LENGTH(cons(N, L)) → A(L)
U12¹(tt, L) → A(L)
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U11¹(tt, L) → A(L)
U12¹(tt, L) → LENGTH(a(L))
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

A(zerosInact) → ZEROS
LENGTH(cons(N, L)) → A(L)
U12¹(tt, L) → A(L)
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U11¹(tt, L) → A(L)
U12¹(tt, L) → LENGTH(a(L))
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(cons(N, L)) → U11¹(tt, a(L))
U11¹(tt, L) → U12¹(tt, a(L))
U12¹(tt, L) → LENGTH(a(L))

The TRS R consists of the following rules:

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ Narrowing

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(cons(N, L)) → U11¹(tt, a(L))
U12¹(tt, L) → LENGTH(a(L))
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U12¹(tt, L) → LENGTH(a(L)) at position [0] we obtained the following new rules:

U12¹(tt, zerosInact) → LENGTH(zeros)
U12¹(tt, x0) → LENGTH(x0)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U12¹(tt, zerosInact) → LENGTH(zeros)
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U12¹(tt, x0) → LENGTH(x0)
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U12¹(tt, zerosInact) → LENGTH(zeros) at position [0] we obtained the following new rules:

U12¹(tt, zerosInact) → LENGTH(zerosInact)
U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ DependencyGraphProof

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))
U12¹(tt, zerosInact) → LENGTH(zerosInact)
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U11¹(tt, L) → U12¹(tt, a(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

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

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ NonTerminationProof

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U12¹(tt, x0) → LENGTH(x0)
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

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 narrowing to the left:

The TRS P consists of the following rules:

U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U12¹(tt, x0) → LENGTH(x0)
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

s = LENGTH(cons(N, zerosInact)) evaluates to t =LENGTH(cons(0, zerosInact))

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

Semiunifier: [ ]
Matcher: [N / 0]

Rewriting sequence

LENGTH(cons(N, zerosInact)) → U11¹(tt, a(zerosInact))
with rule LENGTH(cons(N', L)) → U11¹(tt, a(L)) at position [] and matcher [L / zerosInact, N' / N]

U11¹(tt, a(zerosInact)) → U11¹(tt, zerosInact)
with rule a(x) → x at position [1] and matcher [x / zerosInact]

U11¹(tt, zerosInact) → U12¹(tt, a(zerosInact))
with rule U11¹(tt, L) → U12¹(tt, a(L)) at position [] and matcher [L / zerosInact]

U12¹(tt, a(zerosInact)) → U12¹(tt, zerosInact)
with rule a(x) → x at position [1] and matcher [x / zerosInact]

U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))
with rule U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))

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 Innermost Giesl Middeldorp transformation [10] to transform the context-sensitive TRS to a usual TRS.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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

active(length(nil)) → mark(0)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(U11(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U12(x₁, x₂)) = x₁ + 2·x₂
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 2
POL(s(x₁)) = x₁
POL(tt) = 0
POL(zeros) = 0

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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

CONS(mark(x1), x2) → CONS(x1, x2)
ACTIVE(U11(tt, L)) → U12¹(tt, L)
MARK(tt) → ACTIVE(tt)
MARK(U11(x1, x2)) → MARK(x1)
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(U11(x1, x2)) → U11¹(mark(x1), x2)
ACTIVE(U12(tt, L)) → S(length(L))
MARK(length(x1)) → MARK(x1)
MARK(U12(x1, x2)) → U12¹(mark(x1), x2)
MARK(s(x1)) → ACTIVE(s(mark(x1)))
ACTIVE(zeros) → CONS(0, zeros)
LENGTH(mark(x1)) → LENGTH(x1)
U12¹(active(x1), x2) → U12¹(x1, x2)
LENGTH(active(x1)) → LENGTH(x1)
ACTIVE(U12(tt, L)) → MARK(s(length(L)))
MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
U11¹(active(x1), x2) → U11¹(x1, x2)
S(mark(x1)) → S(x1)
S(active(x1)) → S(x1)
MARK(s(x1)) → S(mark(x1))
MARK(U12(x1, x2)) → MARK(x1)
MARK(cons(x1, x2)) → CONS(mark(x1), x2)
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(cons(x1, x2)) → ACTIVE(cons(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(length(cons(N, L))) → U11¹(tt, L)
MARK(length(x1)) → LENGTH(mark(x1))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(0) → ACTIVE(0)
U12¹(mark(x1), x2) → U12¹(x1, x2)
CONS(active(x1), x2) → CONS(x1, x2)
U11¹(mark(x1), x2) → U11¹(x1, x2)
MARK(nil) → ACTIVE(nil)
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → LENGTH(L)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

CONS(mark(x1), x2) → CONS(x1, x2)
ACTIVE(U11(tt, L)) → U12¹(tt, L)
MARK(tt) → ACTIVE(tt)
MARK(U11(x1, x2)) → MARK(x1)
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(U11(x1, x2)) → U11¹(mark(x1), x2)
ACTIVE(U12(tt, L)) → S(length(L))
MARK(length(x1)) → MARK(x1)
MARK(U12(x1, x2)) → U12¹(mark(x1), x2)
MARK(s(x1)) → ACTIVE(s(mark(x1)))
ACTIVE(zeros) → CONS(0, zeros)
LENGTH(mark(x1)) → LENGTH(x1)
U12¹(active(x1), x2) → U12¹(x1, x2)
LENGTH(active(x1)) → LENGTH(x1)
ACTIVE(U12(tt, L)) → MARK(s(length(L)))
MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
U11¹(active(x1), x2) → U11¹(x1, x2)
S(mark(x1)) → S(x1)
S(active(x1)) → S(x1)
MARK(s(x1)) → S(mark(x1))
MARK(U12(x1, x2)) → MARK(x1)
MARK(cons(x1, x2)) → CONS(mark(x1), x2)
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(cons(x1, x2)) → ACTIVE(cons(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(length(cons(N, L))) → U11¹(tt, L)
MARK(length(x1)) → LENGTH(mark(x1))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(0) → ACTIVE(0)
U12¹(mark(x1), x2) → U12¹(x1, x2)
CONS(active(x1), x2) → CONS(x1, x2)
U11¹(mark(x1), x2) → U11¹(x1, x2)
MARK(nil) → ACTIVE(nil)
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → LENGTH(L)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 13 less nodes.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(mark(x1)) → LENGTH(x1)
LENGTH(active(x1)) → LENGTH(x1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(mark(x1)) → LENGTH(x1)
LENGTH(active(x1)) → LENGTH(x1)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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.

cons(active(x0), x1)
cons(mark(x0), x1)
U11(active(x0), x1)
U11(mark(x0), x1)
U12(active(x0), x1)
U12(mark(x0), x1)
s(active(x0))
s(mark(x0))
length(active(x0))
length(mark(x0))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(mark(x1)) → LENGTH(x1)
LENGTH(active(x1)) → LENGTH(x1)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
mark(0)
mark(U11(x0, x1))
mark(tt)
mark(U12(x0, x1))
mark(s(x0))
mark(length(x0))
mark(nil)

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:

LENGTH(mark(x1)) → LENGTH(x1)
The graph contains the following edges 1 > 1
LENGTH(active(x1)) → LENGTH(x1)
The graph contains the following edges 1 > 1

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

S(active(x1)) → S(x1)
S(mark(x1)) → S(x1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

S(active(x1)) → S(x1)
S(mark(x1)) → S(x1)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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.

cons(active(x0), x1)
cons(mark(x0), x1)
U11(active(x0), x1)
U11(mark(x0), x1)
U12(active(x0), x1)
U12(mark(x0), x1)
s(active(x0))
s(mark(x0))
length(active(x0))
length(mark(x0))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

S(mark(x1)) → S(x1)
S(active(x1)) → S(x1)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
mark(0)
mark(U11(x0, x1))
mark(tt)
mark(U12(x0, x1))
mark(s(x0))
mark(length(x0))
mark(nil)

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:

S(active(x1)) → S(x1)
The graph contains the following edges 1 > 1
S(mark(x1)) → S(x1)
The graph contains the following edges 1 > 1

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U12¹(mark(x1), x2) → U12¹(x1, x2)
U12¹(active(x1), x2) → U12¹(x1, x2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U12¹(mark(x1), x2) → U12¹(x1, x2)
U12¹(active(x1), x2) → U12¹(x1, x2)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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.

cons(active(x0), x1)
cons(mark(x0), x1)
U11(active(x0), x1)
U11(mark(x0), x1)
U12(active(x0), x1)
U12(mark(x0), x1)
s(active(x0))
s(mark(x0))
length(active(x0))
length(mark(x0))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U12¹(mark(x1), x2) → U12¹(x1, x2)
U12¹(active(x1), x2) → U12¹(x1, x2)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
mark(0)
mark(U11(x0, x1))
mark(tt)
mark(U12(x0, x1))
mark(s(x0))
mark(length(x0))
mark(nil)

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:

U12¹(mark(x1), x2) → U12¹(x1, x2)
The graph contains the following edges 1 > 1, 2 >= 2
U12¹(active(x1), x2) → U12¹(x1, x2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U11¹(active(x1), x2) → U11¹(x1, x2)
U11¹(mark(x1), x2) → U11¹(x1, x2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U11¹(active(x1), x2) → U11¹(x1, x2)
U11¹(mark(x1), x2) → U11¹(x1, x2)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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.

cons(active(x0), x1)
cons(mark(x0), x1)
U11(active(x0), x1)
U11(mark(x0), x1)
U12(active(x0), x1)
U12(mark(x0), x1)
s(active(x0))
s(mark(x0))
length(active(x0))
length(mark(x0))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

U11¹(active(x1), x2) → U11¹(x1, x2)
U11¹(mark(x1), x2) → U11¹(x1, x2)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
mark(0)
mark(U11(x0, x1))
mark(tt)
mark(U12(x0, x1))
mark(s(x0))
mark(length(x0))
mark(nil)

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:

U11¹(active(x1), x2) → U11¹(x1, x2)
The graph contains the following edges 1 > 1, 2 >= 2
U11¹(mark(x1), x2) → U11¹(x1, x2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

CONS(mark(x1), x2) → CONS(x1, x2)
CONS(active(x1), x2) → CONS(x1, x2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

CONS(mark(x1), x2) → CONS(x1, x2)
CONS(active(x1), x2) → CONS(x1, x2)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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.

cons(active(x0), x1)
cons(mark(x0), x1)
U11(active(x0), x1)
U11(mark(x0), x1)
U12(active(x0), x1)
U12(mark(x0), x1)
s(active(x0))
s(mark(x0))
length(active(x0))
length(mark(x0))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QReductionProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                      ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

CONS(mark(x1), x2) → CONS(x1, x2)
CONS(active(x1), x2) → CONS(x1, x2)

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

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
mark(0)
mark(U11(x0, x1))
mark(tt)
mark(U12(x0, x1))
mark(s(x0))
mark(length(x0))
mark(nil)

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:

CONS(mark(x1), x2) → CONS(x1, x2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS(active(x1), x2) → CONS(x1, x2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ RuleRemovalProof

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
MARK(U12(x1, x2)) → MARK(x1)
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(U11(x1, x2)) → MARK(x1)
MARK(cons(x1, x2)) → ACTIVE(cons(mark(x1), x2))
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
MARK(length(x1)) → MARK(x1)
MARK(s(x1)) → ACTIVE(s(mark(x1)))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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(U12(x1, x2)) → MARK(x1)
MARK(U11(x1, x2)) → MARK(x1)
MARK(length(x1)) → MARK(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = 2·x₁
POL(MARK(x₁)) = 2·x₁
POL(U11(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(U12(x₁, x₂)) = 1 + x₁ + x₂
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tt) = 0
POL(zeros) = 0

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(zeros) → ACTIVE(zeros)
MARK(s(x1)) → ACTIVE(s(mark(x1)))
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → ACTIVE(cons(mark(x1), x2))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(x1, x2)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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(zeros) → ACTIVE(zeros)

The remaining pairs can at least be oriented weakly.

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(s(x1)) → ACTIVE(s(mark(x1)))
MARK(s(x1)) → MARK(x1)
MARK(cons(x1, x2)) → ACTIVE(cons(mark(x1), x2))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(x1, x2)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

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

POL(0) = 0
POL(ACTIVE(x₁)) = 0
POL(MARK(x₁)) = x₁
POL(U11(x₁, x₂)) = 0
POL(U12(x₁, x₂)) = 0
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁
POL(length(x₁)) = 0
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tt) = 0
POL(zeros) = 1

The following usable rules [17] were oriented:

length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
s(mark(x1)) → s(x1)
s(active(x1)) → s(x1)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(cons(x1, x2)) → ACTIVE(cons(mark(x1), x2))
MARK(s(x1)) → MARK(x1)
MARK(s(x1)) → ACTIVE(s(mark(x1)))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(x1, x2)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(cons(x1, x2)) → ACTIVE(cons(mark(x1), x2))
MARK(s(x1)) → MARK(x1)
MARK(s(x1)) → ACTIVE(s(mark(x1)))
MARK(cons(x1, x2)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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(cons(x1, x2)) → ACTIVE(cons(mark(x1), x2))
MARK(cons(x1, x2)) → MARK(x1)

The remaining pairs can at least be oriented weakly.

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(s(x1)) → MARK(x1)
MARK(s(x1)) → ACTIVE(s(mark(x1)))
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

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

POL(0) = 0
POL(ACTIVE(x₁)) = 0
POL(MARK(x₁)) = x₁
POL(U11(x₁, x₂)) = 0
POL(U12(x₁, x₂)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₁
POL(length(x₁)) = 0
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(tt) = 0
POL(zeros) = 0

The following usable rules [17] were oriented:

length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
s(mark(x1)) → s(x1)
s(active(x1)) → s(x1)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(s(x1)) → ACTIVE(s(mark(x1)))
MARK(s(x1)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(s(x1)) → ACTIVE(s(mark(x1)))

The remaining pairs can at least be oriented weakly.

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(s(x1)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

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

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 1
POL(U11(x₁, x₂)) = 1
POL(U12(x₁, x₂)) = 1
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(length(x₁)) = 1
POL(mark(x₁)) = 0
POL(nil) = 0
POL(s(x₁)) = 0
POL(tt) = 0
POL(zeros) = 0

The following usable rules [17] were oriented:

U11(mark(x1), x2) → U11(x1, x2)
U11(active(x1), x2) → U11(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
U12(active(x1), x2) → U12(x1, x2)
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
s(mark(x1)) → s(x1)
s(active(x1)) → s(x1)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ AND

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

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

MARK(U11(x1, x2)) → ACTIVE(U11(mark(x1), x2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(s(x1)) → MARK(x1)
MARK(U12(x1, x2)) → ACTIVE(U12(mark(x1), x2))
MARK(length(x1)) → ACTIVE(length(mark(x1)))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(x1, x2)) → active(cons(mark(x1), x2))
cons(active(x1), x2) → cons(x1, x2)
cons(mark(x1), x2) → cons(x1, x2)
mark(0) → active(0)
mark(U11(x1, x2)) → active(U11(mark(x1), x2))
U11(active(x1), x2) → U11(x1, x2)
U11(mark(x1), x2) → U11(x1, x2)
mark(tt) → active(tt)
mark(U12(x1, x2)) → active(U12(mark(x1), x2))
U12(active(x1), x2) → U12(x1, x2)
U12(mark(x1), x2) → U12(x1, x2)
mark(s(x1)) → active(s(mark(x1)))
s(active(x1)) → s(x1)
s(mark(x1)) → s(x1)
mark(length(x1)) → active(length(mark(x1)))
length(active(x1)) → length(x1)
length(mark(x1)) → length(x1)
mark(nil) → active(nil)

The set Q consists of the following terms:

active(zeros)
active(U11(tt, x0))
active(U12(tt, x0))
active(length(nil))
active(length(cons(x0, x1)))
mark(zeros)
mark(cons(x0, x1))
cons(active(x0), x1)
cons(mark(x0), x1)
mark(0)
mark(U11(x0, x1))
U11(active(x0), x1)
U11(mark(x0), x1)
mark(tt)
mark(U12(x0, x1))
U12(active(x0), x1)
U12(mark(x0), x1)
mark(s(x0))
s(active(x0))
s(mark(x0))
mark(length(x0))
length(active(x0))
length(mark(x0))
mark(nil)

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

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

      ↳ Improved Ferreira Ribeiro-Transformation

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

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(nil) → 0
length(cons(N, L)) → U11(tt, L)

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)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(nil) → 0
length(cons(N, L)) → U11(tt, L)

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(U11(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(U12(x₁, x₂)) = 2 + x₁ + x₂
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = 2 + x₁
POL(nil) = 1
POL(s(x₁)) = x₁
POL(tt) = 0
POL(zeros) = 0

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

      ↳ Improved Ferreira Ribeiro-Transformation

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

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(cons(N, L)) → U11(tt, L)

Q is empty.

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

                ↳ QTRS

                  ↳ DependencyPairsProof

      ↳ Improved Ferreira Ribeiro-Transformation

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

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(cons(N, L)) → U11(tt, L)

The set Q consists of the following terms:

zeros
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

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

ZEROS → ZEROS
LENGTH(cons(N, L)) → U11¹(tt, L)
U12¹(tt, L) → LENGTH(L)
U11¹(tt, L) → U12¹(tt, L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(cons(N, L)) → U11(tt, L)

The set Q consists of the following terms:

zeros
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

      ↳ Improved Ferreira Ribeiro-Transformation

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

ZEROS → ZEROS
LENGTH(cons(N, L)) → U11¹(tt, L)
U12¹(tt, L) → LENGTH(L)
U11¹(tt, L) → U12¹(tt, L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(cons(N, L)) → U11(tt, L)

The set Q consists of the following terms:

zeros
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(cons(N, L)) → U11¹(tt, L)
U11¹(tt, L) → U12¹(tt, L)
U12¹(tt, L) → LENGTH(L)

The TRS R consists of the following rules:

zeros → cons(0, zeros)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(cons(N, L)) → U11(tt, L)

The set Q consists of the following terms:

zeros
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                          ↳ QDP

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(cons(N, L)) → U11¹(tt, L)
U11¹(tt, L) → U12¹(tt, L)
U12¹(tt, L) → LENGTH(L)

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

zeros
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

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
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ QDPSizeChangeProof

                          ↳ QDP

      ↳ Improved Ferreira Ribeiro-Transformation

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

LENGTH(cons(N, L)) → U11¹(tt, L)
U12¹(tt, L) → LENGTH(L)
U11¹(tt, L) → U12¹(tt, L)

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:

U11¹(tt, L) → U12¹(tt, L)
The graph contains the following edges 1 >= 1, 2 >= 2
U12¹(tt, L) → LENGTH(L)
The graph contains the following edges 2 >= 1
LENGTH(cons(N, L)) → U11¹(tt, L)
The graph contains the following edges 1 > 2

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

      ↳ Improved Ferreira Ribeiro-Transformation

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)
U11(tt, L) → U12(tt, L)
U12(tt, L) → s(length(L))
length(cons(N, L)) → U11(tt, L)

The set Q consists of the following terms:

zeros
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

      ↳ Improved Ferreira Ribeiro-Transformation

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
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

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
U11(tt, x0)
U12(tt, x0)
length(cons(x0, x1))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ Overlay + Local Confluence

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ NonTerminationProof

      ↳ Improved Ferreira Ribeiro-Transformation

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.
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:

ZEROS → ZEROS

The TRS R consists of the following rules:none

s = ZEROS evaluates to t =ZEROS

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

Matcher: [ ]
Semiunifier: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from ZEROS to ZEROS.

We applied the Improved Ferreira Ribeiro transformation [5,11] to transform the context-sensitive TRS to a usual TRS.

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

Q is empty.

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

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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(U11(x₁, x₂)) = x₁ + 2·x₂
POL(U12(x₁, x₂)) = x₁ + 2·x₂
POL(a(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(length(x₁)) = x₁
POL(nil) = 2
POL(s(x₁)) = x₁
POL(tt) = 0
POL(zeros) = 0
POL(zerosInact) = 0

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

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

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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:

A(zerosInact) → ZEROS
LENGTH(cons(N, L)) → A(L)
U12¹(tt, L) → A(L)
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U11¹(tt, L) → A(L)
U12¹(tt, L) → LENGTH(a(L))
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

A(zerosInact) → ZEROS
LENGTH(cons(N, L)) → A(L)
U12¹(tt, L) → A(L)
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U11¹(tt, L) → A(L)
U12¹(tt, L) → LENGTH(a(L))
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

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

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

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

LENGTH(cons(N, L)) → U11¹(tt, a(L))
U11¹(tt, L) → U12¹(tt, a(L))
U12¹(tt, L) → LENGTH(a(L))

The TRS R consists of the following rules:

zeros → cons(0, zerosInact)
U11(tt, L) → U12(tt, a(L))
U12(tt, L) → s(length(a(L)))
length(cons(N, L)) → U11(tt, a(L))
a(x) → x
zeros → zerosInact
a(zerosInact) → zeros

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ Narrowing

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

LENGTH(cons(N, L)) → U11¹(tt, a(L))
U12¹(tt, L) → LENGTH(a(L))
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U12¹(tt, L) → LENGTH(a(L)) at position [0] we obtained the following new rules:

U12¹(tt, zerosInact) → LENGTH(zeros)
U12¹(tt, x0) → LENGTH(x0)

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

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

U12¹(tt, zerosInact) → LENGTH(zeros)
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U12¹(tt, x0) → LENGTH(x0)
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U12¹(tt, zerosInact) → LENGTH(zeros) at position [0] we obtained the following new rules:

U12¹(tt, zerosInact) → LENGTH(zerosInact)
U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))

↳ CSR

  ↳ CSRInnermostProof

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ DependencyGraphProof

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

U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))
U12¹(tt, zerosInact) → LENGTH(zerosInact)
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U11¹(tt, L) → U12¹(tt, a(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

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

    ↳ CSR

      ↳ CSDependencyPairsProof

      ↳ Zantema-Transformation

      ↳ Innermost Giesl Middeldorp-Transformation

      ↳ Trivial-Transformation

      ↳ Improved Ferreira Ribeiro-Transformation

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ DependencyPairsProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ Narrowing

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ NonTerminationProof

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

U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U12¹(tt, x0) → LENGTH(x0)
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

U12¹(tt, zerosInact) → LENGTH(cons(0, zerosInact))
LENGTH(cons(N, L)) → U11¹(tt, a(L))
U12¹(tt, x0) → LENGTH(x0)
U11¹(tt, L) → U12¹(tt, a(L))

The TRS R consists of the following rules:

a(x) → x
a(zerosInact) → zeros
zeros → cons(0, zerosInact)
zeros → zerosInact

s = LENGTH(cons(N, zerosInact)) evaluates to t =LENGTH(cons(0, zerosInact))

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

Matcher: [N / 0]
Semiunifier: [ ]