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))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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))) → F(g(f(a)))
ACTIVE(f(f(a))) → G(f(a))
ACTIVE(f(X)) → F(active(X))
ACTIVE(f(X)) → ACTIVE(X)
F(mark(X)) → F(X)
PROPER(f(X)) → F(proper(X))
PROPER(f(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
F(ok(X)) → F(X)
G(ok(X)) → G(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 5 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(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
ok(x1)  =  ok(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

F(ok(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))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(ok(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
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) 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(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(g(X)) → PROPER(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

g₁: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(f(X)) → PROPER(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

f₁: [1]

The following usable rules [FROCOS05] were oriented: none

(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))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

ACTIVE(f(X)) → ACTIVE(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

f₁: [1]

The following usable rules [FROCOS05] were oriented: none

(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))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

(29) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(mark(X)) → TOP(proper(X))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[a, g] > [mark₁, f₁]

Status:

f₁: [1]
a: []
g: []
mark₁: [1]
TOP₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(31) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(ok(X)) → TOP(active(X))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[TOP₁, ok] > active > a > mark₁
[TOP₁, ok] > active > a > g₁

Status:

active: []
a: []
g₁: [1]
mark₁: [1]
TOP₁: [1]
ok: []

The following usable rules [FROCOS05] were oriented:

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

(33) Obligation:

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

active(f(f(a))) → mark(f(g(f(a))))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(f(X)) → f(proper(X))
proper(a) → ok(a)
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE