Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

active(h(X)) → mark(c(d(X)))

Used ordering:
Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

H(active(X)) → H(X)
ACTIVE(f(f(X))) → C(f(g(f(X))))
C(active(X)) → C(X)
MARK(c(X)) → ACTIVE(c(X))
C(mark(X)) → C(X)
D(active(X)) → D(X)
MARK(f(X)) → MARK(X)
F(active(X)) → F(X)
MARK(h(X)) → ACTIVE(h(mark(X)))
F(mark(X)) → F(X)
ACTIVE(c(X)) → D(X)
ACTIVE(c(X)) → MARK(d(X))
ACTIVE(f(f(X))) → G(f(X))
ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
G(active(X)) → G(X)
D(mark(X)) → D(X)
G(mark(X)) → G(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(h(X)) → MARK(X)
MARK(d(X)) → ACTIVE(d(X))
MARK(g(X)) → ACTIVE(g(X))
MARK(f(X)) → F(mark(X))
H(mark(X)) → H(X)
ACTIVE(f(f(X))) → F(g(f(X)))
MARK(h(X)) → H(mark(X))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

H(active(X)) → H(X)
ACTIVE(f(f(X))) → C(f(g(f(X))))
C(active(X)) → C(X)
MARK(c(X)) → ACTIVE(c(X))
C(mark(X)) → C(X)
D(active(X)) → D(X)
MARK(f(X)) → MARK(X)
F(active(X)) → F(X)
MARK(h(X)) → ACTIVE(h(mark(X)))
F(mark(X)) → F(X)
ACTIVE(c(X)) → D(X)
ACTIVE(c(X)) → MARK(d(X))
ACTIVE(f(f(X))) → G(f(X))
ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
G(active(X)) → G(X)
D(mark(X)) → D(X)
G(mark(X)) → G(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(h(X)) → MARK(X)
MARK(d(X)) → ACTIVE(d(X))
MARK(g(X)) → ACTIVE(g(X))
MARK(f(X)) → F(mark(X))
H(mark(X)) → H(X)
ACTIVE(f(f(X))) → F(g(f(X)))
MARK(h(X)) → H(mark(X))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 6 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

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

                    ↳ UsableRulesReductionPairsProof

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

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:

H(active(X)) → H(X)
H(mark(X)) → H(X)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(H(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.
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

                ↳ UsableRulesProof

                  ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

D(mark(X)) → D(X)
D(active(X)) → D(X)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

D(mark(X)) → D(X)
D(active(X)) → D(X)

R is empty.
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

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

D(mark(X)) → D(X)
D(active(X)) → D(X)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(D(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(G(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(active(X)) → C(X)
C(mark(X)) → C(X)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(active(X)) → C(X)
C(mark(X)) → C(X)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(C(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

              ↳ QDP

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(active(X)) → C(X)
C(mark(X)) → C(X)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x₁)) = 2·x₁
POL(active(x₁)) = 2·x₁
POL(mark(x₁)) = 2·x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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(h(X)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁
POL(f(x₁)) = x₁
POL(g(x₁)) = x₁
POL(h(x₁)) = 2 + x₁
POL(mark(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

MARK(f(X)) → MARK(X)

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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(h(X)) → ACTIVE(h(mark(X)))
MARK(g(X)) → ACTIVE(g(X))

The remaining pairs can at least be oriented weakly.

ACTIVE(c(X)) → MARK(d(X))
MARK(d(X)) → ACTIVE(d(X))
ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(c(X)) → ACTIVE(c(X))
MARK(f(X)) → ACTIVE(f(mark(X)))

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( active(x₁) ) = 1

POL( f(x₁) ) = max{0, -1}

POL( c(x₁) ) = max{0, -1}

POL( MARK(x₁) ) = x₁ + 1

POL( g(x₁) ) = 1

POL( h(x₁) ) = x₁ + 1

POL( mark(x₁) ) = 0

POL( d(x₁) ) = max{0, -1}

POL( ACTIVE(x₁) ) = 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

ACTIVE(c(X)) → MARK(d(X))
MARK(d(X)) → ACTIVE(d(X))
ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(c(X)) → ACTIVE(c(X))
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

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

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(d(X)) → ACTIVE(d(X))

The remaining pairs can at least be oriented weakly.

ACTIVE(c(X)) → MARK(d(X))
ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(c(X)) → ACTIVE(c(X))
MARK(f(X)) → ACTIVE(f(mark(X)))

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( active(x₁) ) = max{0, -1}

POL( f(x₁) ) = 1

POL( c(x₁) ) = 1

POL( MARK(x₁) ) = 1

POL( g(x₁) ) = max{0, -1}

POL( h(x₁) ) = x₁ + 1

POL( mark(x₁) ) = max{0, -1}

POL( d(x₁) ) = max{0, -1}

POL( ACTIVE(x₁) ) = x₁

The following usable rules [17] were oriented:

d(active(X)) → d(X)
d(mark(X)) → d(X)
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(active(X)) → c(X)
c(mark(X)) → c(X)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

ACTIVE(c(X)) → MARK(d(X))
ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(c(X)) → ACTIVE(c(X))
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

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

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(f(X)) → ACTIVE(f(mark(X)))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( active(x₁) ) = max{0, -1}

POL( f(x₁) ) = 1

POL( c(x₁) ) = max{0, -1}

POL( MARK(x₁) ) = x₁ + 1

POL( g(x₁) ) = max{0, -1}

POL( h(x₁) ) = max{0, x₁ - 1}

POL( mark(x₁) ) = max{0, x₁ - 1}

POL( d(x₁) ) = max{0, -1}

POL( ACTIVE(x₁) ) = 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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.

No dependency pairs are removed.

The following rules are removed from R:

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

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = 2·x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁
POL(f(x₁)) = 1 + 2·x₁
POL(g(x₁)) = x₁
POL(mark(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                                          ↳ QDP

                                            ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

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 0 SCCs with 3 less nodes.
We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

f(f(active(x))) → f(g(f(c(mark(x)))))
c(active(x)) → d(mark(x))
h(active(x)) → d(c(mark(x)))
f(mark(x)) → mark(f(active(x)))
c(mark(x)) → c(active(x))
g(mark(x)) → g(active(x))
d(mark(x)) → d(active(x))
h(mark(x)) → mark(h(active(x)))
mark(f(x)) → f(x)
active(f(x)) → f(x)
mark(c(x)) → c(x)
active(c(x)) → c(x)
mark(g(x)) → g(x)
active(g(x)) → g(x)
mark(d(x)) → d(x)
active(d(x)) → d(x)
mark(h(x)) → h(x)
active(h(x)) → h(x)

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

f(f(active(x))) → f(g(f(c(mark(x)))))
c(active(x)) → d(mark(x))
h(active(x)) → d(c(mark(x)))
f(mark(x)) → mark(f(active(x)))
c(mark(x)) → c(active(x))
g(mark(x)) → g(active(x))
d(mark(x)) → d(active(x))
h(mark(x)) → mark(h(active(x)))
mark(f(x)) → f(x)
active(f(x)) → f(x)
mark(c(x)) → c(x)
active(c(x)) → c(x)
mark(g(x)) → g(x)
active(g(x)) → g(x)
mark(d(x)) → d(x)
active(d(x)) → d(x)
mark(h(x)) → h(x)
active(h(x)) → h(x)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

f(f(active(x))) → f(g(f(c(mark(x)))))
c(active(x)) → d(mark(x))
h(active(x)) → d(c(mark(x)))
f(mark(x)) → mark(f(active(x)))
c(mark(x)) → c(active(x))
g(mark(x)) → g(active(x))
d(mark(x)) → d(active(x))
h(mark(x)) → mark(h(active(x)))
mark(f(x)) → f(x)
active(f(x)) → f(x)
mark(c(x)) → c(x)
active(c(x)) → c(x)
mark(g(x)) → g(x)
active(g(x)) → g(x)
mark(d(x)) → d(x)
active(d(x)) → d(x)
mark(h(x)) → h(x)
active(h(x)) → h(x)

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

f(f(active(x))) → f(g(f(c(mark(x)))))
c(active(x)) → d(mark(x))
h(active(x)) → d(c(mark(x)))
f(mark(x)) → mark(f(active(x)))
c(mark(x)) → c(active(x))
g(mark(x)) → g(active(x))
d(mark(x)) → d(active(x))
h(mark(x)) → mark(h(active(x)))
mark(f(x)) → f(x)
active(f(x)) → f(x)
mark(c(x)) → c(x)
active(c(x)) → c(x)
mark(g(x)) → g(x)
active(g(x)) → g(x)
mark(d(x)) → d(x)
active(d(x)) → d(x)
mark(h(x)) → h(x)
active(h(x)) → h(x)

Q is empty.