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 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 DependencyGraphProof (⇔)
    ↳26 TRUE

(0) Obligation:

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

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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(a, X, X)) → MARK(f(X, b, b))
ACTIVE(f(a, X, X)) → F(X, b, b)
ACTIVE(b) → MARK(a)
MARK(f(X1, X2, X3)) → ACTIVE(f(X1, mark(X2), X3))
MARK(f(X1, X2, X3)) → F(X1, mark(X2), X3)
MARK(f(X1, X2, X3)) → MARK(X2)
MARK(a) → ACTIVE(a)
MARK(b) → ACTIVE(b)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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 2 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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.


F(X1, X2, mark(X3)) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[F1, mark1]

Status:
mark1: multiset
F1: multiset


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

F(X1, mark(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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.


F(X1, mark(X2), X3) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x2, x3)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > F2

Status:
mark1: [1]
F2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

F(mark(X1), X2, X3) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(mark(X1), X2, X3) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x1, x2, x3)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[F3, mark1]

Status:
mark1: [1]
F3: multiset


The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(X1, X2, active(X3)) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
active(x1)  =  active(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
[F1, active1]

Status:
active1: multiset
F1: [1]


The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

F(active(X1), X2, X3) → F(X1, X2, X3)
F(X1, active(X2), X3) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(active(X1), X2, X3) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x1, x3)
active(x1)  =  active(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
[F2, active1]

Status:
active1: multiset
F2: multiset


The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

F(X1, active(X2), X3) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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.


F(X1, active(X2), X3) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x2, x3)
active(x1)  =  active(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
[F2, active1]

Status:
active1: multiset
F2: [1,2]


The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, mark(X2), X3))
ACTIVE(f(a, X, X)) → MARK(f(X, b, b))
MARK(f(X1, X2, X3)) → MARK(X2)

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(f(X1, X2, X3)) → MARK(X2)
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, x3)  =  f(x1, x2)
ACTIVE(x1)  =  ACTIVE(x1)
mark(x1)  =  x1
a  =  a
b  =  b
active(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
f2 > [MARK1, ACTIVE1] > [a, b]

Status:
MARK1: [1]
a: multiset
f2: multiset
b: multiset
ACTIVE1: [1]


The following usable rules [FROCOS05] were oriented:

mark(b) → active(b)
mark(a) → active(a)
f(X1, X2, active(X3)) → f(X1, X2, X3)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
active(f(a, X, X)) → mark(f(X, b, b))
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
active(b) → mark(a)

(22) Obligation:

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, mark(X2), X3))
ACTIVE(f(a, X, X)) → MARK(f(X, b, b))

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(f(a, X, X)) → MARK(f(X, b, b))
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, x3)  =  f(x1, x3)
ACTIVE(x1)  =  ACTIVE(x1)
mark(x1)  =  x1
a  =  a
b  =  b
active(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
f2 > [MARK1, ACTIVE1] > b
a > [MARK1, ACTIVE1] > b

Status:
MARK1: multiset
a: multiset
f2: multiset
b: multiset
ACTIVE1: multiset


The following usable rules [FROCOS05] were oriented:

f(X1, X2, active(X3)) → f(X1, X2, X3)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)

(24) Obligation:

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, mark(X2), X3))

The TRS R consists of the following rules:

active(f(a, X, X)) → mark(f(X, b, b))
active(b) → mark(a)
mark(f(X1, X2, X3)) → active(f(X1, mark(X2), X3))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(26) TRUE