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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

MARK(h(X)) → ACTIVE(h(mark(X)))
F(mark(X)) → F(X)
H(active(X)) → H(X)
ACTIVE(f(X)) → H(f(X))
G(active(X)) → G(X)
G(mark(X)) → G(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(h(X)) → MARK(X)
ACTIVE(f(X)) → G(h(f(X)))
MARK(f(X)) → F(mark(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(f(X)) → MARK(g(h(f(X))))
H(mark(X)) → H(X)
MARK(h(X)) → H(mark(X))
MARK(f(X)) → MARK(X)
F(active(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

MARK(h(X)) → ACTIVE(h(mark(X)))
F(mark(X)) → F(X)
H(active(X)) → H(X)
ACTIVE(f(X)) → H(f(X))
G(active(X)) → G(X)
G(mark(X)) → G(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(h(X)) → MARK(X)
ACTIVE(f(X)) → G(h(f(X)))
MARK(f(X)) → F(mark(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(f(X)) → MARK(g(h(f(X))))
H(mark(X)) → H(X)
MARK(h(X)) → H(mark(X))
MARK(f(X)) → MARK(X)
F(active(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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 4 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
          ↳ 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(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
            ↳ UsableRulesProof
          ↳ 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.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
          ↳ 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
          ↳ 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(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
            ↳ UsableRulesProof
          ↳ 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
          ↳ 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
            ↳ 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.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                    ↳ PisEmptyProof
            ↳ UsableRulesProof
          ↳ 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
            ↳ UsableRulesProof
              ↳ QDP
          ↳ 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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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)))
MARK(h(X)) → MARK(X)
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(f(X)) → MARK(X)
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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)) = 2·x1   
POL(MARK(x1)) = 2·x1   
POL(active(x1)) = x1   
POL(f(x1)) = 1 + 2·x1   
POL(g(x1)) = x1   
POL(h(x1)) = x1   
POL(mark(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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)))
MARK(h(X)) → MARK(X)
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(f(X)) → ACTIVE(f(mark(X)))

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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(h(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(g(X)) → ACTIVE(g(X))
ACTIVE(f(X)) → MARK(g(h(f(X))))
MARK(f(X)) → ACTIVE(f(mark(X)))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( active(x1) ) = x1


POL( MARK(x1) ) = x1


POL( f(x1) ) = x1


POL( g(x1) ) = max{0, -1}


POL( h(x1) ) = x1 + 1


POL( mark(x1) ) = 0


POL( ACTIVE(x1) ) = max{0, -1}



The following usable rules [17] were oriented:

g(active(X)) → g(X)
g(mark(X)) → g(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ RuleRemovalProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

MARK(g(X)) → ACTIVE(g(X))
ACTIVE(f(X)) → MARK(g(h(f(X))))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( active(x1) ) = max{0, x1 - 1}


POL( MARK(x1) ) = x1 + 1


POL( f(x1) ) = x1 + 1


POL( g(x1) ) = max{0, -1}


POL( h(x1) ) = 0


POL( mark(x1) ) = 1


POL( ACTIVE(x1) ) = 1



The following usable rules [17] were oriented:

g(active(X)) → g(X)
g(mark(X)) → g(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ RuleRemovalProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ UsableRulesProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
mark(f(X)) → active(f(mark(X)))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ RuleRemovalProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ UsableRulesReductionPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

g(active(X)) → g(X)
g(mark(X)) → g(X)
f(mark(X)) → f(X)
f(active(X)) → f(X)
h(active(X)) → h(X)
h(mark(X)) → h(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)
h(active(X)) → h(X)
h(mark(X)) → h(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(h(x1)) = x1   
POL(mark(x1)) = 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ RuleRemovalProof
              ↳ 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:

MARK(g(X)) → ACTIVE(g(X))
ACTIVE(f(X)) → MARK(g(h(f(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 2 less nodes.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS

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

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

Q is empty.