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

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) 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(f(g(f(a))))
mark(f(X)) → active(f(X))
mark(a) → active(a)
mark(g(X)) → active(g(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

G(mark(X)) → G(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
f(x1)  =  f(x1)
a  =  a
g(x1)  =  g

Recursive path order with status [RPO].
Precedence:

f₁ > g > mark₁
a > g > mark₁

Status:

mark₁: multiset
f₁: multiset
a: multiset
g: multiset

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

G(active(X)) → G(X)

The TRS R consists of the following rules:

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

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

G(active(X)) → G(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1)  =  G(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
a  =  a
mark(x1)  =  mark(x1)
g(x1)  =  g

Recursive path order with status [RPO].
Precedence:

G₁ > active₁
f₁ > a > active₁
f₁ > mark₁ > active₁
f₁ > g > active₁

Status:

G₁: multiset
active₁: multiset
f₁: [1]
a: multiset
mark₁: multiset
g: multiset

The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(X))
mark(a) → active(a)
mark(g(X)) → active(g(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(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) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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(f(g(f(a))))
mark(f(X)) → active(f(X))
mark(a) → active(a)
mark(g(X)) → active(g(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(mark(X)) → F(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
f(x1)  =  f(x1)
a  =  a
g(x1)  =  g

Recursive path order with status [RPO].
Precedence:

f₁ > g > mark₁
a > g > mark₁

Status:

mark₁: multiset
f₁: multiset
a: multiset
g: multiset

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

F(active(X)) → F(X)

The TRS R consists of the following rules:

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

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(active(X)) → F(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
a  =  a
mark(x1)  =  mark(x1)
g(x1)  =  g

Recursive path order with status [RPO].
Precedence:

F₁ > active₁
f₁ > a > active₁
f₁ > mark₁ > active₁
f₁ > g > active₁

Status:

F₁: multiset
active₁: multiset
f₁: [1]
a: multiset
mark₁: multiset
g: multiset

The following usable rules [FROCOS05] were oriented:

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

(16) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
f(x1)  =  f(x1)
ACTIVE(x1)  =  x1
a  =  a
g(x1)  =  g
active(x1)  =  x1
mark(x1)  =  mark(x1)

Recursive path order with status [RPO].
Precedence:

f₁ > MARK₁
f₁ > g > mark₁
a > MARK₁
a > g > mark₁

Status:

MARK₁: multiset
f₁: multiset
a: multiset
g: multiset
mark₁: multiset

The following usable rules [FROCOS05] were oriented:

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

(21) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

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

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(g(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
g(x1)  =  g(x1)
active(x1)  =  x1
f(x1)  =  f
a  =  a
mark(x1)  =  x1

Recursive path order with status [RPO].
Precedence:

f > a

Status:

MARK₁: [1]
g₁: multiset
f: []
a: multiset

The following usable rules [FROCOS05] were oriented:

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

(26) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE