Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

active(f(f(a))) → mark(f(g(f(a))))
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(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(g(X)) → G(active(X))
ACTIVE(g(X)) → ACTIVE(X)
G(mark(X)) → G(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(g(X)) → g(active(X))
g(mark(X)) → mark(g(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:

F(ok(X)) → F(X)

The TRS R consists of the following rules:

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

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

Recursive path order with status [RPO].
Precedence:

F₁ > ok₁
proper₁ > a > ok₁
top > ok₁

Status:

F₁: multiset
ok₁: [1]
a: multiset
proper₁: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(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(g(X)) → g(active(X))
g(mark(X)) → mark(g(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:

G(ok(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))))
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(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.

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

Recursive path order with status [RPO].
Precedence:

active₁ > mark₁ > G₁
active₁ > a
proper > a

Status:

G₁: multiset
mark₁: multiset
active₁: [1]
a: multiset
proper: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(12) 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(g(X)) → g(active(X))
g(mark(X)) → mark(g(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.

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

Recursive path order with status [RPO].
Precedence:

G₁ > ok₁
proper₁ > a > ok₁
top > ok₁

Status:

G₁: multiset
ok₁: [1]
a: multiset
proper₁: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

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

Recursive path order with status [RPO].
Precedence:

active₁ > f₁ > ok₁ > top
a > f₁ > ok₁ > top
proper₁ > f₁ > ok₁ > top

Status:

PROPER₁: [1]
f₁: multiset
active₁: [1]
a: multiset
proper₁: [1]
ok₁: multiset
top: multiset

The following usable rules [FROCOS05] were oriented:

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

(19) Obligation:

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

PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

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

Recursive path order with status [RPO].
Precedence:

proper₁ > a > g₁ > ok₁
proper₁ > a > f₁ > ok₁
top > active₁ > g₁ > ok₁
top > active₁ > f₁ > ok₁

Status:

g₁: [1]
active₁: multiset
f₁: [1]
a: multiset
proper₁: multiset
ok₁: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(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(g(X)) → g(active(X))
g(mark(X)) → mark(g(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(g(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

Recursive path order with status [RPO].
Precedence:

proper₁ > a > g₁ > ok₁
proper₁ > a > f₁ > ok₁
top > active₁ > g₁ > ok₁
top > active₁ > f₁ > ok₁

Status:

g₁: [1]
active₁: multiset
f₁: [1]
a: multiset
proper₁: multiset
ok₁: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(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(g(X)) → g(active(X))
g(mark(X)) → mark(g(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(g(X)) → g(active(X))
g(mark(X)) → mark(g(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.