Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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

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, 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)  =  x2
mark(x1)  =  mark(x1)
active(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

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

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(mark(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)  =  F(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:

mark₁ > F₂

Status:

mark₁: [1]
F₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

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(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
active(x1)  =  active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

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) → F(X1, X2)

The TRS R consists of the following rules:

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

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(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
active(x1)  =  active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

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

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) Obligation:

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

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

The TRS R consists of the following rules:

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

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.

MARK(f(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
f(x1, x2)  =  f(x1)
ACTIVE(x1)  =  x1
mark(x1)  =  x1
a  =  a
b  =  b
active(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:

f₁ > a
b > a

Status:

f₁: [1]
a: []
b: []

The following usable rules [FROCOS05] were oriented:

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

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

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