Termination w.r.t. Q proof of /tmp/tmpJnSEFK/Ex4_7_77_Bor03

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

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

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(tail(cons(X, XS))) → mark(XS)
mark(tail(X)) → active(tail(mark(X)))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(mark(x₁)) = 2·x₁
POL(tail(x₁)) = 1 + x₁
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

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

active(zeros) → mark(cons(0, zeros))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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:

CONS(mark(X1), X2) → CONS(X1, X2)
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
TAIL(active(X)) → TAIL(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
TAIL(mark(X)) → TAIL(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(0) → ACTIVE(0)
ACTIVE(zeros) → CONS(0, zeros)
CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

CONS(mark(X1), X2) → CONS(X1, X2)
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
TAIL(active(X)) → TAIL(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
TAIL(mark(X)) → TAIL(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(0) → ACTIVE(0)
ACTIVE(zeros) → CONS(0, zeros)
CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 4 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(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:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)

R is empty.
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.

ACTIVE(zeros) → MARK(cons(0, zeros))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = 1
POL(MARK(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁
POL(zeros) = 1

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

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

R is empty.
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

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPSizeChangeProof

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

MARK(cons(X1, X2)) → MARK(X1)

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

MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1