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))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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)) → MARK(h(X))
ACTIVE(g(X)) → H(X)
ACTIVE(c) → MARK(d)
ACTIVE(h(d)) → MARK(g(c))
ACTIVE(h(d)) → G(c)
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(X))
MARK(c) → ACTIVE(c)
MARK(d) → ACTIVE(d)
G(mark(X)) → G(X)
G(active(X)) → G(X)
H(mark(X)) → H(X)
H(active(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))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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 3 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

H(active(X)) → H(X)
H(mark(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))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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(active(X)) → H(X)
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)  =  H(x1)
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
g(x1)  =  g
h(x1)  =  h
c  =  c
d  =  d

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

H₁ > [g, h, d]
c > [active₁, mark₁] > [g, h, d]

Status:

H₁: [1]
active₁: [1]
mark₁: [1]
g: []
h: []
c: []
d: []

The following usable rules [FROCOS05] were oriented:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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(active(X)) → G(X)
G(mark(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))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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(active(X)) → G(X)
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)
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)
g(x1)  =  g
h(x1)  =  h
c  =  c
d  =  d

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

G₁ > [g, h, d]
c > [active₁, mark₁] > [g, h, d]

Status:

G₁: [1]
active₁: [1]
mark₁: [1]
g: []
h: []
c: []
d: []

The following usable rules [FROCOS05] were oriented:

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

MARK(g(X)) → ACTIVE(g(X))
ACTIVE(g(X)) → MARK(h(X))
MARK(h(X)) → ACTIVE(h(X))
ACTIVE(h(d)) → MARK(g(c))

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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