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

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

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

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

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

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

a__and(tt, X) → mark(X)
a__length(nil) → 0
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(tt) → tt
mark(nil) → nil

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

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

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

a__zeros → zeros
a__and(X1, X2) → and(X1, X2)
a__length(X) → length(X)

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

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

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

mark(length(X)) → a__length(mark(X))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__length(x₁)) = x₁
POL(a__zeros) = 0
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(length(x₁)) = 1 + 2·x₁
POL(mark(x₁)) = x₁
POL(s(x₁)) = 2·x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))

Q is empty.

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))

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

mark(0) → 0

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__length(x₁)) = 2·x₁
POL(a__zeros) = 2
POL(cons(x₁, x₂)) = 2 + x₁ + x₂
POL(mark(x₁)) = 2 + x₁
POL(s(x₁)) = x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))

Q is empty.

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))

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

mark(cons(X1, X2)) → cons(mark(X1), X2)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__length(x₁)) = 2·x₁
POL(a__zeros) = 2
POL(cons(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(mark(x₁)) = 2 + 2·x₁
POL(s(x₁)) = x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

Q is empty.

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

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

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
A__LENGTH(cons(N, L)) → MARK(L)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

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

MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
A__LENGTH(cons(N, L)) → A__LENGTH(mark(L))
A__LENGTH(cons(N, L)) → MARK(L)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                  ↳ QDP

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

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                  ↳ QDP

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

MARK(s(X)) → MARK(X)

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

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ QDPSizeChangeProof

                                  ↳ QDP

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

MARK(s(X)) → MARK(X)

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

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

MARK(s(X)) → MARK(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                    ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__length(cons(N, L)) → s(a__length(mark(L)))
mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                    ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

a__length(cons(x0, x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                    ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

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

Strictly oriented rules of the TRS R:

mark(s(X)) → s(mark(X))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__LENGTH(x₁)) = 2·x₁
POL(a__zeros) = 0
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(mark(x₁)) = 2·x₁
POL(s(x₁)) = 1 + x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ Narrowing

                                    ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

mark(zeros) → a__zeros
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                    ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

mark(zeros) → a__zeros
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                    ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

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

mark(zeros)
mark(s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                                          ↳ QDP

                                                            ↳ Rewriting

                                    ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                    ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                    ↳ UsableRulesProof

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

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

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

a__zeros

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.

a__zeros

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                    ↳ UsableRulesProof

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

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

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ UsableRulesProof

                                                      ↳ QDP

                                                        ↳ QReductionProof

                                                          ↳ QDP

                                                            ↳ Rewriting

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ NonTerminationProof

                                    ↳ UsableRulesProof

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

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

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

The TRS P consists of the following rules:

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

The TRS R consists of the following rules:none

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

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

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

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

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

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

The TRS R consists of the following rules:

mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
a__length(cons(x0, x1))
mark(zeros)
mark(s(x0))

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

a__length(cons(x0, x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ MNOCProof

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

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

The TRS R consists of the following rules:

mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))
a__zeros → cons(0, zeros)

The set Q consists of the following terms:

a__zeros
mark(zeros)
mark(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ RRRPoloQTRSProof

                    ↳ QTRS

                      ↳ Overlay + Local Confluence

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ MNOCProof

                                              ↳ QDP

                                                ↳ NonTerminationProof

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

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

The TRS R consists of the following rules:

mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))
a__zeros → cons(0, zeros)

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:

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

The TRS R consists of the following rules:

mark(zeros) → a__zeros
mark(s(X)) → s(mark(X))
a__zeros → cons(0, zeros)

s = A__LENGTH(mark(zeros)) evaluates to t =A__LENGTH(mark(zeros))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

A__LENGTH(mark(zeros)) → A__LENGTH(a__zeros)
with rule mark(zeros) → a__zeros at position [0] and matcher [ ]

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

A__LENGTH(cons(0, zeros)) → A__LENGTH(mark(zeros))
with rule A__LENGTH(cons(N, L)) → A__LENGTH(mark(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.