Termination w.r.t. Q proof of /tmp/tmpZKnoDM/Ex4_7_15_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:

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

Q is empty.

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(f(X)) → a__f(mark(X))
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(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(f(X)) → a__f(mark(X))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__f(x₁)) = 1 + 2·x₁
POL(a__p(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(f(x₁)) = 1 + 2·x₁
POL(mark(x₁)) = 2·x₁
POL(p(x₁)) = x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(X)

Q is empty.

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__f(X) → f(X)
a__p(X) → p(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)
a__f(X) → f(X)

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__f(x₁)) = 1 + x₁
POL(a__p(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(f(x₁)) = x₁
POL(mark(x₁)) = 2·x₁
POL(p(x₁)) = 2·x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
a__p(X) → p(X)

Q is empty.

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

a__f(0) → cons(0, f(s(0)))
a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
a__p(X) → p(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__f(0) → cons(0, f(s(0)))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__f(x₁)) = 2 + 2·x₁
POL(a__p(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(f(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(p(x₁)) = 2·x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

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

a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
a__p(X) → p(X)

Q is empty.

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

a__f(s(0)) → a__f(a__p(s(0)))
a__p(s(0)) → 0
mark(p(X)) → a__p(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
a__p(X) → p(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__p(s(0)) → 0
mark(0) → 0
mark(s(X)) → s(mark(X))

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(a__f(x₁)) = 2·x₁
POL(a__p(x₁)) = x₁
POL(mark(x₁)) = 1 + 2·x₁
POL(p(x₁)) = x₁
POL(s(x₁)) = 2 + 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ AAECC Innermost

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

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The TRS R 2 is

a__f(s(0)) → a__f(a__p(s(0)))

The signature Sigma is {a__f}

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ AAECC Innermost

                    ↳ QTRS

                      ↳ DependencyPairsProof

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

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The set Q consists of the following terms:

a__f(s(0))
mark(p(x0))
a__p(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(p(X)) → A__P(mark(X))
A__F(s(0)) → A__P(s(0))
A__F(s(0)) → A__F(a__p(s(0)))
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The set Q consists of the following terms:

a__f(s(0))
mark(p(x0))
a__p(x0)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ AAECC Innermost

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

MARK(p(X)) → A__P(mark(X))
A__F(s(0)) → A__P(s(0))
A__F(s(0)) → A__F(a__p(s(0)))
MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The set Q consists of the following terms:

a__f(s(0))
mark(p(x0))
a__p(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ AAECC Innermost

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

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

MARK(p(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(s(0)) → a__f(a__p(s(0)))
mark(p(X)) → a__p(mark(X))
a__p(X) → p(X)

The set Q consists of the following terms:

a__f(s(0))
mark(p(x0))
a__p(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

                  ↳ AAECC Innermost

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

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

MARK(p(X)) → MARK(X)

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

a__f(s(0))
mark(p(x0))
a__p(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__f(s(0))
mark(p(x0))
a__p(x0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ AAECC Innermost

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ QDPSizeChangeProof

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

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