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(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
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:
F(mark(X)) → F(X)
ACTIVE(f(f(a))) → G(f(a))
C(active(X)) → C(X)
MARK(g(X)) → ACTIVE(g(mark(X)))
G(active(X)) → G(X)
MARK(c(X)) → ACTIVE(c(X))
G(mark(X)) → G(X)
ACTIVE(f(f(a))) → F(g(f(a)))
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(g(X)) → G(mark(X))
MARK(g(X)) → MARK(X)
MARK(a) → ACTIVE(a)
ACTIVE(f(f(a))) → MARK(c(f(g(f(a)))))
MARK(f(X)) → F(mark(X))
C(mark(X)) → C(X)
ACTIVE(f(f(a))) → C(f(g(f(a))))
F(active(X)) → F(X)
MARK(f(X)) → MARK(X)
The TRS R consists of the following rules:
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
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:
F(mark(X)) → F(X)
ACTIVE(f(f(a))) → G(f(a))
C(active(X)) → C(X)
MARK(g(X)) → ACTIVE(g(mark(X)))
G(active(X)) → G(X)
MARK(c(X)) → ACTIVE(c(X))
G(mark(X)) → G(X)
ACTIVE(f(f(a))) → F(g(f(a)))
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(g(X)) → G(mark(X))
MARK(g(X)) → MARK(X)
MARK(a) → ACTIVE(a)
ACTIVE(f(f(a))) → MARK(c(f(g(f(a)))))
MARK(f(X)) → F(mark(X))
C(mark(X)) → C(X)
ACTIVE(f(f(a))) → C(f(g(f(a))))
F(active(X)) → F(X)
MARK(f(X)) → MARK(X)
The TRS R consists of the following rules:
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 8 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ 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(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
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
↳ 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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ 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)
The TRS R consists of the following rules:
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
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
↳ 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.
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
↳ 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ 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
↳ 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)
The TRS R consists of the following rules:
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
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
↳ 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
↳ 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
↳ 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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(f(a))) → MARK(c(f(g(f(a)))))
MARK(c(X)) → ACTIVE(c(X))
The TRS R consists of the following rules:
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
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
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(f(a))) → MARK(c(f(g(f(a)))))
MARK(c(X)) → ACTIVE(c(X))
The TRS R consists of the following rules:
c(active(X)) → c(X)
c(mark(X)) → c(X)
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 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
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(f(f(a))) → MARK(c(f(g(f(a)))))
MARK(c(X)) → ACTIVE(c(X))
The TRS R consists of the following rules:
c(active(X)) → c(X)
c(mark(X)) → c(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X)) → MARK(X)
MARK(g(X)) → MARK(X)
The TRS R consists of the following rules:
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
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
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
MARK(g(X)) → MARK(X)
MARK(f(X)) → MARK(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
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
MARK(g(X)) → MARK(X)
MARK(f(X)) → MARK(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:
MARK(g(X)) → MARK(X)
MARK(f(X)) → MARK(X)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(MARK(x1)) = 2·x1
POL(f(x1)) = 2·x1
POL(g(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ 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 have reversed the following QTRS:
The set of rules R is
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
The set Q is empty.
We have obtained the following QTRS:
a'(f(f(active(x)))) → a'(f(g(f(c(mark(x))))))
f(mark(x)) → mark(f(active(x)))
a'(mark(x)) → a'(active(x))
c(mark(x)) → c(active(x))
g(mark(x)) → mark(g(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)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a'(f(f(active(x)))) → a'(f(g(f(c(mark(x))))))
f(mark(x)) → mark(f(active(x)))
a'(mark(x)) → a'(active(x))
c(mark(x)) → c(active(x))
g(mark(x)) → mark(g(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)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
active(f(f(a))) → mark(c(f(g(f(a)))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(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)
The set Q is empty.
We have obtained the following QTRS:
a'(f(f(active(x)))) → a'(f(g(f(c(mark(x))))))
f(mark(x)) → mark(f(active(x)))
a'(mark(x)) → a'(active(x))
c(mark(x)) → c(active(x))
g(mark(x)) → mark(g(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)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a'(f(f(active(x)))) → a'(f(g(f(c(mark(x))))))
f(mark(x)) → mark(f(active(x)))
a'(mark(x)) → a'(active(x))
c(mark(x)) → c(active(x))
g(mark(x)) → mark(g(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)
Q is empty.