Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
  ↳1 QTRS Reverse (⇔)
    ↳2 QTRS
  ↳3 QTRS Reverse (⇔)
    ↳4 QTRS
  ↳5 DependencyPairsProof (⇔)
    ↳6 QDP
    ↳7 DependencyGraphProof (⇔)
    ↳8 AND
      ↳9 QDP
        ↳10 UsableRulesProof (⇔)
          ↳11 QDP
          ↳12 UsableRulesReductionPairsProof (⇔)
          ↳13 QDP
          ↳14 PisEmptyProof (⇔)
          ↳15 TRUE
        ↳16 UsableRulesProof (⇔)
          ↳17 QDP
      ↳18 QDP
        ↳19 UsableRulesProof (⇔)
          ↳20 QDP
          ↳21 UsableRulesReductionPairsProof (⇔)
          ↳22 QDP
          ↳23 PisEmptyProof (⇔)
          ↳24 TRUE
        ↳25 UsableRulesProof (⇔)
          ↳26 QDP
      ↳27 QDP
        ↳28 UsableRulesProof (⇔)
          ↳29 QDP
          ↳30 UsableRulesReductionPairsProof (⇔)
          ↳31 QDP
          ↳32 PisEmptyProof (⇔)
          ↳33 TRUE
        ↳34 UsableRulesProof (⇔)
          ↳35 QDP
      ↳36 QDP
        ↳37 UsableRulesProof (⇔)
          ↳38 QDP
          ↳39 DependencyGraphProof (⇔)
          ↳40 TRUE
        ↳41 UsableRulesProof (⇔)
          ↳42 QDP
      ↳43 QDP
        ↳44 UsableRulesProof (⇔)
          ↳45 QDP
          ↳46 UsableRulesReductionPairsProof (⇔)
          ↳47 QDP
          ↳48 PisEmptyProof (⇔)
          ↳49 TRUE
        ↳50 UsableRulesProof (⇔)
          ↳51 QDP

(0) Obligation:

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.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

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.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

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

ACTIVE(f(f(a))) → MARK(c(f(g(f(a)))))
ACTIVE(f(f(a))) → C(f(g(f(a))))
ACTIVE(f(f(a))) → F(g(f(a)))
ACTIVE(f(f(a))) → G(f(a))
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → F(mark(X))
MARK(f(X)) → MARK(X)
MARK(a) → ACTIVE(a)
MARK(c(X)) → ACTIVE(c(X))
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → G(mark(X))
MARK(g(X)) → 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 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.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 8 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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.

(10) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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.

(12) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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 [POLO]:

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

(13) Obligation:

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

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) TRUE

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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.

(18) Obligation:

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.

(19) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(20) Obligation:

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.

(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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 [POLO]:

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

(22) Obligation:

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(24) TRUE

(25) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(26) Obligation:

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.

(27) Obligation:

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

F(active(X)) → F(X)
F(mark(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.

(28) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(29) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

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

(30) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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(active(X)) → F(X)
F(mark(X)) → F(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

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

(31) Obligation:

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

(32) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(33) TRUE

(34) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(35) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

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

(36) Obligation:

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

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

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.

(37) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(38) Obligation:

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

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

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.

(39) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(40) TRUE

(41) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(42) Obligation:

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

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

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.

(43) Obligation:

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

MARK(g(X)) → MARK(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.

(44) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(45) Obligation:

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.

(46) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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 [POLO]:

POL(MARK(x1)) = 2·x1   
POL(f(x1)) = 2·x1   
POL(g(x1)) = 2·x1   

(47) Obligation:

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

(48) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(49) TRUE

(50) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(51) Obligation:

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.