(0) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(g(X), Y)) → F(X, f(g(X), Y))
ACTIVE(f(X1, X2)) → F(active(X1), X2)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
ACTIVE(g(X)) → G(active(X))
ACTIVE(g(X)) → ACTIVE(X)
F(mark(X1), X2) → F(X1, X2)
G(mark(X)) → G(X)
PROPER(f(X1, X2)) → F(proper(X1), proper(X2))
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(f(X1, X2)) → PROPER(X2)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
F(ok(X1), ok(X2)) → F(X1, X2)
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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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)
mark(x1)  =  x1
active(x1)  =  active(x1)
f(x1, x2)  =  f(x2)
g(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > f1 > ok1

Status:
ok1: [1]
active1: [1]
f1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) 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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) 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
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1, x2)  =  x1
g(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > mark1
ok1 > mark1
top > mark1

Status:
mark1: [1]
active1: [1]
ok1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(9) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(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(ok(X1), ok(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(ok(X1), ok(X2)) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x2)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  x1
f(x1, x2)  =  f(x1)
g(x1)  =  g(x1)
proper(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
f1 > ok1
g1 > ok1

Status:
F1: [1]
ok1: [1]
f1: [1]
g1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(14) Obligation:

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

F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(mark(X1), X2) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1, x2)  =  x1
g(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
F2 > mark1
active1 > mark1
ok1 > mark1
top > mark1

Status:
F2: [1,2]
mark1: [1]
active1: [1]
ok1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(16) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(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:

PROPER(f(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
f(x1, x2)  =  f(x1, x2)
g(x1)  =  x1
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > f2 > ok > mark1
top > mark1

Status:
f2: [1,2]
active1: [1]
mark1: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(21) 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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) 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)  =  PROPER(x1)
g(x1)  =  g(x1)
active(x1)  =  x1
f(x1, x2)  =  f
mark(x1)  =  mark
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
f > g1 > PROPER1 > mark
top > mark

Status:
PROPER1: [1]
g1: [1]
f: []
mark: []
top: []

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(23) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(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:

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(f(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) 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)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
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)
f(x1, x2)  =  f(x1)
active(x1)  =  x1
mark(x1)  =  mark(x1)
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
f1 > g1 > mark1
f1 > ok

Status:
g1: [1]
f1: [1]
mark1: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(28) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) TRUE

(31) 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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  x1
mark(x1)  =  x1
proper(x1)  =  x1
f(x1, x2)  =  x1
g(x1)  =  x1
top(x1)  =  top(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]
top1: [1]

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(33) Obligation:

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

TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) 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)
mark(x1)  =  mark(x1)
proper(x1)  =  x1
active(x1)  =  active(x1)
f(x1, x2)  =  f(x1)
g(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
ok > active1 > f1 > mark1 > TOP1
top > active1 > f1 > mark1 > TOP1

Status:
TOP1: [1]
mark1: [1]
active1: [1]
f1: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(35) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE