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(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.

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

(3) DependencyGraphProof (EQUIVALENT transformation)

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

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

H(active(X)) → H(X)

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

Recursive Path Order [RPO].
Precedence:

[H₁, active₁]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

H(mark(X)) → H(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:

mark₁ > H₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
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.

(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:

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.

(13) 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)
mark(x1)  =  x1

Recursive Path Order [RPO].
Precedence:

[G₁, active₁]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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.

(15) 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: Recursive Path Order [RPO].
Precedence:

mark₁ > G₁

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

Q DP problem:
P is empty.
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.

(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:

F(active(X)) → F(X)
F(mark(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.

(20) 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)
mark(x1)  =  x1

Recursive Path Order [RPO].
Precedence:

[F₁, active₁]

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

F(mark(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.

(22) 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: Recursive Path Order [RPO].
Precedence:

mark₁ > F₁

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

Q DP problem:
P is empty.
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.

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(h(X)) → ACTIVE(h(mark(X)))

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

Recursive Path Order [RPO].
Precedence:

active > [MARK, f, mark, g] > h

The following usable rules [FROCOS05] were oriented:

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

(28) Obligation:

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

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(g(X)) → ACTIVE(g(X))

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

Recursive Path Order [RPO].
Precedence:

[MARK, f, active] > mark
[MARK, f, active] > g

The following usable rules [FROCOS05] were oriented:

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

(30) Obligation:

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

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(f(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)
f(x1)  =  f(x1)
ACTIVE(x1)  =  x1
mark(x1)  =  mark(x1)
g(x1)  =  g
h(x1)  =  x1
active(x1)  =  x1

Recursive Path Order [RPO].
Precedence:

[MARK₁, f₁, mark₁] > g

The following usable rules [FROCOS05] were oriented:

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

(32) Obligation:

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

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(h(X)) → MARK(X)

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

Recursive Path Order [RPO].
Precedence:

active > [f, ACTIVE, g, h₁] > mark

The following usable rules [FROCOS05] were oriented:

g(active(X)) → g(X)
g(mark(X)) → g(X)

(34) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(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.

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

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)  =  ACTIVE
mark(x1)  =  mark
g(x1)  =  g
h(x1)  =  h(x1)
active(x1)  =  active

Recursive Path Order [RPO].
Precedence:

[f₁, ACTIVE, active] > g
[f₁, ACTIVE, active] > h₁ > [MARK₁, mark]

The following usable rules [FROCOS05] were oriented:

g(active(X)) → g(X)
g(mark(X)) → g(X)

(36) Obligation:

Q DP problem:
P is empty.
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.

(37) PisEmptyProof (EQUIVALENT transformation)

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

(38) TRUE