Termination w.r.t. Q proof of /tmp/tmpCKimbq/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:

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

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

active(and(tt, X)) → mark(X)

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

active(length(nil)) → mark(0)

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

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

Q is empty.

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

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

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

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.

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(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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(tt) → active(tt)

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

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

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

Q is empty.

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

ACTIVE(length(cons(N, L))) → S(length(L))
CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
ACTIVE(length(cons(N, L))) → LENGTH(L)
AND(X1, mark(X2)) → AND(X1, X2)
MARK(length(X)) → MARK(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → CONS(0, zeros)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(length(X)) → ACTIVE(length(mark(X)))
AND(X1, active(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
MARK(zeros) → ACTIVE(zeros)
MARK(length(X)) → LENGTH(mark(X))
MARK(0) → ACTIVE(0)
ACTIVE(zeros) → MARK(cons(0, zeros))
CONS(active(X1), X2) → CONS(X1, X2)
AND(active(X1), X2) → AND(X1, X2)

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

ACTIVE(length(cons(N, L))) → S(length(L))
CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
ACTIVE(length(cons(N, L))) → LENGTH(L)
AND(X1, mark(X2)) → AND(X1, X2)
MARK(length(X)) → MARK(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → CONS(0, zeros)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(length(X)) → ACTIVE(length(mark(X)))
AND(X1, active(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
MARK(zeros) → ACTIVE(zeros)
MARK(length(X)) → LENGTH(mark(X))
MARK(0) → ACTIVE(0)
ACTIVE(zeros) → MARK(cons(0, zeros))
CONS(active(X1), X2) → CONS(X1, X2)
AND(active(X1), X2) → AND(X1, X2)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 7 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

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

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.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

S(active(X)) → S(X)
S(mark(X)) → S(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:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

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

AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

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

AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

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

AND(mark(X1), X2) → AND(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
AND(active(X1), X2) → AND(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
AND(X1, mark(X2)) → AND(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
AND(X1, active(X2)) → AND(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

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

CONS(X1, active(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

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

CONS(X1, active(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
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
CONS(X1, mark(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

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

MARK(length(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

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

MARK(length(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))

The TRS R consists of the following rules:

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

Q is empty.
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(length(X)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))

The TRS R consists of the following rules:

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

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

The following pairs can be oriented strictly and are deleted.

MARK(zeros) → ACTIVE(zeros)

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))

Used ordering: Polynomial interpretation [25]:

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

The following usable rules [17] were oriented:

length(active(X)) → length(X)
length(mark(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))

The TRS R consists of the following rules:

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

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(X)) → ACTIVE(length(mark(X)))

Used ordering: Polynomial interpretation [25]:

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

The following usable rules [17] were oriented:

length(active(X)) → length(X)
length(mark(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

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

ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(mark(X)))

The TRS R consists of the following rules:

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

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

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.

ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(mark(X)))

Used ordering: Polynomial interpretation [25]:

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

The following usable rules [17] were oriented:

length(active(X)) → length(X)
length(mark(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

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

MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(X)) → ACTIVE(length(mark(X)))

The TRS R consists of the following rules:

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

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

MARK(length(0)) → ACTIVE(length(active(0)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ Narrowing

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

MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(0)) → ACTIVE(length(active(0)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(length(0)) → ACTIVE(length(active(0))) at position [0] we obtained the following new rules:

MARK(length(0)) → ACTIVE(length(0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

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

MARK(length(0)) → ACTIVE(length(0))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ RuleRemovalProof

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

MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))

The TRS R consists of the following rules:

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

MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ RuleRemovalProof

                                                                  ↳ QDP

                                                                    ↳ QDPOrderProof

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

MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))

The TRS R consists of the following rules:

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

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

The following pairs can be oriented strictly and are deleted.

MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))

The remaining pairs can at least be oriented weakly.

MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(x0)) → ACTIVE(length(x0))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( active(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	1	1	\
\	0	0	/

x₂

M( zeros ) =

/	0	\
\	0	/

M( mark(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( s(x₁) ) =

/	0	\
\	1	/

/	1	0	\
\	0	0	/

x₁

M( length(x₁) ) =

/	0	\
\	1	/

/	1	1	\
\	1	0	/

x₁

M( 0 ) =

/	0	\
\	0	/

Tuple symbols:

M( MARK(x₁) ) =

[

]

x₁

M( ACTIVE(x₁) ) =

[

]

x₁

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ RuleRemovalProof

                                                                  ↳ QDP

                                                                    ↳ QDPOrderProof

                                                                      ↳ QDP

                                                                        ↳ NonTerminationProof

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

MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))

The TRS R consists of the following rules:

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

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:

MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))

The TRS R consists of the following rules:

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

s = ACTIVE(length(active(zeros))) evaluates to t =ACTIVE(length(active(zeros)))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

ACTIVE(length(active(zeros))) → ACTIVE(length(mark(cons(0, zeros))))
with rule active(zeros) → mark(cons(0, zeros)) at position [0,0] and matcher [ ]

ACTIVE(length(mark(cons(0, zeros)))) → ACTIVE(length(cons(0, zeros)))
with rule length(mark(X)) → length(X) at position [0] and matcher [X / cons(0, zeros)]

ACTIVE(length(cons(0, zeros))) → MARK(s(length(zeros)))
with rule ACTIVE(length(cons(N, L))) → MARK(s(length(L))) at position [] and matcher [N / 0, L / zeros]

MARK(s(length(zeros))) → MARK(length(zeros))
with rule MARK(s(X)) → MARK(X) at position [] and matcher [X / length(zeros)]

MARK(length(zeros)) → ACTIVE(length(active(zeros)))
with rule MARK(length(zeros)) → ACTIVE(length(active(zeros)))

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.