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(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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(a))) → MARK(f(g(f(a))))
ACTIVE(f(f(a))) → F(g(f(a)))
ACTIVE(f(f(a))) → G(f(a))
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → F(mark(X))
MARK(f(X)) → MARK(X)
MARK(a) → ACTIVE(a)
MARK(g(X)) → ACTIVE(g(X))
F(mark(X)) → F(X)
F(active(X)) → F(X)
G(mark(X)) → G(X)
G(active(X)) → G(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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 4 less nodes.

(4) Complex Obligation (AND)

(5) 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(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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(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

(7) Obligation:

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

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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:

F(active(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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.

F(active(X)) → F(X)
F(mark(X)) → F(X)

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

POL(F(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(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(f(a))) → MARK(f(g(f(a))))
MARK(f(X)) → MARK(X)
MARK(g(X)) → ACTIVE(g(X))

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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.

MARK(g(X)) → ACTIVE(g(X))

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

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

The following usable rules [FROCOS05] were oriented:

f(active(X)) → f(X)
f(mark(X)) → f(X)
g(active(X)) → g(X)
g(mark(X)) → g(X)

(17) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(f(a))) → MARK(f(g(f(a))))
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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.

ACTIVE(f(f(a))) → MARK(f(g(f(a))))

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

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

The following usable rules [FROCOS05] were oriented:

mark(f(X)) → active(f(mark(X)))
active(f(f(a))) → mark(f(g(f(a))))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(active(X)) → f(X)
f(mark(X)) → f(X)
g(active(X)) → g(X)
g(mark(X)) → g(X)

(19) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(20) DependencyGraphProof (EQUIVALENT transformation)

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

(21) Obligation:

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

MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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.

MARK(f(X)) → MARK(X)

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

POL(MARK(x₁)) = x₁   
POL(f(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

active(f(f(a))) → mark(f(g(f(a))))
mark(f(X)) → active(f(mark(X)))
mark(a) → active(a)
mark(g(X)) → active(g(X))
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(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