Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 PisEmptyProof (⇔)
    ↳23 TRUE

(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(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)  =  x1
mark(x1)  =  mark(x1)
f(x1)  =  x1
a  =  a
g(x1)  =  g

Recursive Path Order [RPO].
Precedence:
a > mark1 > G1
a > g > G1

The following usable rules [FROCOS05] were oriented:

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)

(7) Obligation:

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

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.

(8) 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)
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)
f(x1)  =  f(x1)
a  =  a
mark(x1)  =  mark(x1)
g(x1)  =  g

Recursive Path Order [RPO].
Precedence:
G1 > f1
a > mark1 > g > active1 > f1

The following usable rules [FROCOS05] were oriented:

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)

(9) 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.

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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.

(13) 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)  =  F(x1)
active(x1)  =  x1
mark(x1)  =  mark(x1)
f(x1)  =  x1
a  =  a
g(x1)  =  g

Recursive Path Order [RPO].
Precedence:
a > mark1 > F1
a > g > F1

The following usable rules [FROCOS05] were oriented:

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)

(14) Obligation:

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

F(active(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.

(15) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
a  =  a
mark(x1)  =  mark(x1)
g(x1)  =  g

Recursive Path Order [RPO].
Precedence:
F1 > f1
a > mark1 > g > active1 > f1

The following usable rules [FROCOS05] were oriented:

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)

(16) 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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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 remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
f(x1)  =  f(x1)
ACTIVE(x1)  =  x1
mark(x1)  =  x1
a  =  a
g(x1)  =  g
active(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
f1 > a > MARK1 > g

The following usable rules [FROCOS05] were oriented:

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)

(21) 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.

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE