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)

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
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)

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:

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)

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
              ↳ 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)

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
              ↳ 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)

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:

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)

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
              ↳ 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)

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:

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

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
              ↳ 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)

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
              ↳ 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)

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
              ↳ 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)

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:

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)

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(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)

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
              ↳ 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.