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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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(c) → MARK(f(g(c)))
ACTIVE(c) → F(g(c))
ACTIVE(c) → G(c)
ACTIVE(f(g(X))) → MARK(g(X))
MARK(c) → ACTIVE(c)
MARK(f(X)) → ACTIVE(f(X))
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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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)
c  =  c
f(x1)  =  f
g(x1)  =  g

Lexicographic path order with status [LPO].
Precedence:
c > f > g > mark1 > G1

Status:
G1: [1]
mark1: [1]
c: []
f: []
g: []

The following usable rules [FROCOS05] were oriented:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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)
c  =  c
mark(x1)  =  mark(x1)
f(x1)  =  f
g(x1)  =  g

Lexicographic path order with status [LPO].
Precedence:
G1 > active1
c > f > mark1 > active1
c > f > g > active1

Status:
G1: [1]
active1: [1]
c: []
mark1: [1]
f: []
g: []

The following usable rules [FROCOS05] were oriented:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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)
c  =  c
f(x1)  =  f
g(x1)  =  g

Lexicographic path order with status [LPO].
Precedence:
c > f > g > mark1 > F1

Status:
F1: [1]
mark1: [1]
c: []
f: []
g: []

The following usable rules [FROCOS05] were oriented:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
mark(c) → active(c)
mark(f(X)) → active(f(X))
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)
c  =  c
mark(x1)  =  mark(x1)
f(x1)  =  f
g(x1)  =  g

Lexicographic path order with status [LPO].
Precedence:
F1 > active1
c > f > mark1 > active1
c > f > g > active1

Status:
F1: [1]
active1: [1]
c: []
mark1: [1]
f: []
g: []

The following usable rules [FROCOS05] were oriented:

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

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

The TRS R consists of the following rules:

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


ACTIVE(f(g(X))) → MARK(g(X))
MARK(f(X)) → ACTIVE(f(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.
ACTIVE(x1)  =  x1
f(x1)  =  f(x1)
g(x1)  =  g
MARK(x1)  =  MARK(x1)
active(x1)  =  x1
c  =  c
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
c > f1 > MARK1
c > g > MARK1

Status:
f1: [1]
g: []
MARK1: [1]
c: []

The following usable rules [FROCOS05] were oriented:

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