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(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: Polynomial interpretation [POLO]:

POL(H(x₁)) = x₁   
POL(active(x₁)) = 1 + x₁   
POL(mark(x₁)) = 1 + x₁   

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(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: Polynomial interpretation [POLO]:

POL(G(x₁)) = x₁   
POL(active(x₁)) = 1 + x₁   
POL(mark(x₁)) = 1 + x₁   

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(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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(h(d)) → MARK(g(c))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x₁)) = x₁   
POL(MARK(x₁)) = x₁   
POL(active(x₁)) = x₁   
POL(c) = 0   
POL(d) = 1   
POL(g(x₁)) = x₁   
POL(h(x₁)) = x₁   
POL(mark(x₁)) = x₁   

The following usable rules [FROCOS05] were oriented:

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

(17) 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))

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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(g(X)) → ACTIVE(g(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x₁)) = 0   
POL(MARK(x₁)) = x₁   
POL(active(x₁)) = 0   
POL(g(x₁)) = 1   
POL(h(x₁)) = 0   
POL(mark(x₁)) = 0   

The following usable rules [FROCOS05] were oriented:

h(active(X)) → h(X)
h(mark(X)) → h(X)

(19) Obligation:

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

ACTIVE(g(X)) → MARK(h(X))
MARK(h(X)) → ACTIVE(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.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(g(X)) → MARK(h(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x₁)) = x₁   
POL(MARK(x₁)) = 0   
POL(active(x₁)) = 0   
POL(g(x₁)) = 1   
POL(h(x₁)) = 0   
POL(mark(x₁)) = 0   

The following usable rules [FROCOS05] were oriented:

h(active(X)) → h(X)
h(mark(X)) → h(X)

(21) Obligation:

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

MARK(h(X)) → ACTIVE(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.

(22) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(23) TRUE