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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(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(g(X)) → H(X)
ACTIVE(h(d)) → G(c)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → H(proper(X))
PROPER(h(X)) → PROPER(X)
G(ok(X)) → G(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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(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 4 SCCs with 6 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

H(ok(X)) → H(X)

The TRS R consists of the following rules:

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

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)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
g(x1)  =  x1
mark(x1)  =  mark
h(x1)  =  h(x1)
c  =  c
d  =  d
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:

[ok₁, active₁, h₁, proper₁] > [mark, c, d, top]

Status:

ok₁: multiset
active₁: multiset
mark: multiset
h₁: multiset
c: multiset
d: multiset
proper₁: multiset
top: []

The following usable rules [FROCOS05] were oriented:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
g(ok(X)) → ok(g(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)
active(x1)  =  active(x1)
g(x1)  =  x1
mark(x1)  =  mark
h(x1)  =  h(x1)
c  =  c
d  =  d
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:

[ok₁, active₁, h₁, proper₁] > [mark, c, d, top]

Status:

ok₁: multiset
active₁: multiset
mark: multiset
h₁: multiset
c: multiset
d: multiset
proper₁: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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

PROPER(h(X)) → PROPER(X)
PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

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

PROPER(h(X)) → PROPER(X)
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)
h(x1)  =  h(x1)
g(x1)  =  g(x1)
active(x1)  =  active(x1)
mark(x1)  =  x1
c  =  c
d  =  d
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:

PROPER₁ > [g₁, c, ok, top]
[h₁, active₁, d] > [g₁, c, ok, top]

Status:

PROPER₁: multiset
h₁: multiset
g₁: multiset
active₁: multiset
c: multiset
d: multiset
ok: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(17) Obligation:

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

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

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

The TRS R consists of the following rules:

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