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 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(g(X), Y)) → MARK(f(X, f(g(X), Y)))
ACTIVE(f(g(X), Y)) → F(X, f(g(X), Y))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(f(X1, X2)) → F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → G(mark(X))
MARK(g(X)) → MARK(X)
F(mark(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
G(mark(X)) → G(X)
G(active(X)) → G(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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 3 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(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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)
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)  =  x1

Lexicographic Path Order [LPO].
Precedence:
active1 > G1

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(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)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(X1, mark(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(X1, mark(X2)) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
mark1 > F1

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
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(X1, active(X2)) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x2
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(active(X1), X2) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x1
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(mark(X1), X2) → F(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → MARK(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, x2)  =  f(x1)
ACTIVE(x1)  =  ACTIVE(x1)
mark(x1)  =  x1
g(x1)  =  g(x1)
active(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
f1 > MARK1 > ACTIVE1
f1 > g1 > ACTIVE1

The following usable rules [FROCOS05] were oriented:

g(active(X)) → g(X)
g(mark(X)) → g(X)
mark(g(X)) → active(g(mark(X)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)

(25) Obligation:

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE