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(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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(X))) → C(f(g(f(X))))
ACTIVE(f(f(X))) → F(g(f(X)))
ACTIVE(f(f(X))) → G(f(X))
ACTIVE(c(X)) → D(X)
ACTIVE(h(X)) → C(d(X))
ACTIVE(h(X)) → D(X)
ACTIVE(f(X)) → F(active(X))
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(h(X)) → H(active(X))
ACTIVE(h(X)) → ACTIVE(X)
F(mark(X)) → F(X)
H(mark(X)) → H(X)
PROPER(f(X)) → F(proper(X))
PROPER(f(X)) → PROPER(X)
PROPER(c(X)) → C(proper(X))
PROPER(c(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → D(proper(X))
PROPER(d(X)) → PROPER(X)
PROPER(h(X)) → H(proper(X))
PROPER(h(X)) → PROPER(X)
F(ok(X)) → F(X)
C(ok(X)) → C(X)
G(ok(X)) → G(X)
D(ok(X)) → D(X)
H(ok(X)) → H(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(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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 8 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

D(ok(X)) → D(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

D(ok(X)) → D(X)

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

Lexicographic path order with status [LPO].
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(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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(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].
Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

C(ok(X)) → C(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

C(ok(X)) → C(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

H(ok(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

H(ok(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)
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:

ok₁ > H₁

Status:

H₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(22) 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(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(23) 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: Combined order from the following AFS and order.
H(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) 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(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(28) 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)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:

ok₁ > F₁

Status:

ok₁: [1]
F₁: [1]

The following usable rules [FROCOS05] were oriented: none

(29) 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(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

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].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

PROPER(c(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(h(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)
c(x1)  =  x1
f(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1
h(x1)  =  h(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

PROPER₁: [1]
h₁: [1]

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

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

PROPER(c(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(c(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)
c(x1)  =  c(x1)
f(x1)  =  x1
g(x1)  =  x1
d(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:

c₁ > PROPER₁

Status:

PROPER₁: [1]
c₁: [1]

The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

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

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

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(d(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)
f(x1)  =  x1
g(x1)  =  x1
d(x1)  =  d(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

PROPER₁: [1]
d₁: [1]

The following usable rules [FROCOS05] were oriented: none

(40) Obligation:

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

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

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(41) 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)
f(x1)  =  f(x1)
g(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:

f₁ > PROPER₁

Status:

PROPER₁: [1]
f₁: [1]

The following usable rules [FROCOS05] were oriented: none

(42) 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(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(43) 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)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

g₁: [1]

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(h(X)) → ACTIVE(X)

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

Lexicographic path order with status [LPO].
Precedence:

h₁ > ACTIVE₁

Status:

h₁: [1]
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(49) 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(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(50) 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].
Precedence:

trivial

Status:

f₁: [1]

The following usable rules [FROCOS05] were oriented: none

(51) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(52) PisEmptyProof (EQUIVALENT transformation)

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

(53) TRUE

(54) 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(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(55) 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(x1)
active(x1)  =  active(x1)
mark(x1)  =  x1
proper(x1)  =  x1
c(x1)  =  x1
f(x1)  =  x1
d(x1)  =  x1
g(x1)  =  x1
h(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:

ok₁ > TOP₁ > active₁

Status:

active₁: [1]
ok₁: [1]
TOP₁: [1]

The following usable rules [FROCOS05] were oriented:

c(ok(X)) → ok(c(X))
f(ok(X)) → ok(f(X))
d(ok(X)) → ok(d(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))

(56) 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(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(57) 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)  =  proper(x1)
h(x1)  =  x1
f(x1)  =  x1
d(x1)  =  x1
c(x1)  =  x1
g(x1)  =  g(x1)
ok(x1)  =  ok

Lexicographic path order with status [LPO].
Precedence:

mark₁ > proper₁ > g₁ > TOP₁
ok > g₁ > TOP₁

Status:

g₁: [1]
mark₁: [1]
TOP₁: [1]
ok: []
proper₁: [1]

The following usable rules [FROCOS05] were oriented:

h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
f(mark(X)) → mark(f(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
c(ok(X)) → ok(c(X))
f(ok(X)) → ok(f(X))
d(ok(X)) → ok(d(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))

(58) Obligation:

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(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.

(59) PisEmptyProof (EQUIVALENT transformation)

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

(60) TRUE