Termination w.r.t. Q proof of /tmp/tmpsxAA-f/Ex1_Zan97

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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

Q is empty.

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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:

g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

Used ordering:
Polynomial interpretation [25]:

POL(active(x₁)) = 1 + 2·x₁
POL(c) = 1
POL(d) = 1
POL(g(x₁)) = 1 + x₁
POL(h(x₁)) = 1 + x₁
POL(mark(x₁)) = 1 + 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ AAECC Innermost

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)

The signature Sigma is {active, mark}

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ AAECC Innermost

        ↳ QTRS

          ↳ DependencyPairsProof

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)

The set Q consists of the following terms:

active(g(x0))
active(c)
active(h(d))
mark(g(x0))
mark(h(x0))
mark(c)
mark(d)

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(d) → ACTIVE(d)
MARK(c) → ACTIVE(c)
MARK(h(X)) → ACTIVE(h(X))
ACTIVE(c) → MARK(d)
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(g(X)) → MARK(h(X))
ACTIVE(h(d)) → MARK(g(c))

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)

The set Q consists of the following terms:

active(g(x0))
active(c)
active(h(d))
mark(g(x0))
mark(h(x0))
mark(c)
mark(d)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ AAECC Innermost

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MARK(d) → ACTIVE(d)
MARK(c) → ACTIVE(c)
MARK(h(X)) → ACTIVE(h(X))
ACTIVE(c) → MARK(d)
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(g(X)) → MARK(h(X))
ACTIVE(h(d)) → MARK(g(c))

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)

The set Q consists of the following terms:

active(g(x0))
active(c)
active(h(d))
mark(g(x0))
mark(h(x0))
mark(c)
mark(d)

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

      ↳ AAECC Innermost

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                  ↳ UsableRulesProof

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

MARK(h(X)) → ACTIVE(h(X))
ACTIVE(g(X)) → MARK(h(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(h(d)) → MARK(g(c))

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)

The set Q consists of the following terms:

active(g(x0))
active(c)
active(h(d))
mark(g(x0))
mark(h(x0))
mark(c)
mark(d)

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

      ↳ AAECC Innermost

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

                  ↳ UsableRulesProof

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

MARK(h(X)) → ACTIVE(h(X))
ACTIVE(g(X)) → MARK(h(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(h(d)) → MARK(g(c))

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

active(g(x0))
active(c)
active(h(d))
mark(g(x0))
mark(h(x0))
mark(c)
mark(d)

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.

active(g(x0))
active(c)
active(h(d))
mark(g(x0))
mark(h(x0))
mark(c)
mark(d)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ AAECC Innermost

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

                        ↳ QDP

                  ↳ UsableRulesProof

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

MARK(h(X)) → ACTIVE(h(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(g(X)) → MARK(h(X))
ACTIVE(h(d)) → MARK(g(c))

R is empty.
Q is empty.
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

      ↳ AAECC Innermost

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

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

MARK(h(X)) → ACTIVE(h(X))
ACTIVE(g(X)) → MARK(h(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(h(d)) → MARK(g(c))

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

active(g(x0))
active(c)
active(h(d))
mark(g(x0))
mark(h(x0))
mark(c)
mark(d)

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.

active(g(x0))
active(c)
active(h(d))
mark(g(x0))
mark(h(x0))
mark(c)
mark(d)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ AAECC Innermost

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

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

MARK(h(X)) → ACTIVE(h(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(g(X)) → MARK(h(X))
ACTIVE(h(d)) → MARK(g(c))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

MARK(h(X)) → ACTIVE(h(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(g(X)) → MARK(h(X))
ACTIVE(h(d)) → MARK(g(c))

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = 1 + 2·x₁
POL(MARK(x₁)) = 2 + 2·x₁
POL(c) = 0
POL(d) = 2
POL(g(x₁)) = 1 + 2·x₁
POL(h(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ AAECC Innermost

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.