Termination w.r.t. Q proof of /tmp/tmpXJEgcZ/Ex9_Luc06

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

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 PisEmptyProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

(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)
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)
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1, x2, x3) = f
a = a
b = b

Lexicographic path order with status [LPO].
Quasi-Precedence:

f > b > [mark₁, active₁] > F₁
f > b > [mark₁, active₁] > a

Status:

F₁: [1]
mark₁: [1]
active₁: [1]
f: []
a: []
b: []

The following usable rules [FROCOS05] were oriented:

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)

(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)

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)
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)
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1, x2, x3) = f
a = a
b = b

Lexicographic path order with status [LPO].
Quasi-Precedence:

F₁ > [mark₁, active₁]
f > [a, b] > [mark₁, active₁]

Status:

F₁: [1]
mark₁: [1]
active₁: [1]
f: []
a: []
b: []

The following usable rules [FROCOS05] were oriented:

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)

(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)

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)
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)
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1, x2, x3) = f
a = a
b = b

Lexicographic path order with status [LPO].
Quasi-Precedence:

F₂ > [f, a]
[mark₁, active₁, b] > [f, a]

Status:

F₂: [2,1]
mark₁: [1]
active₁: [1]
f: []
a: []
b: []

The following usable rules [FROCOS05] were oriented:

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)

(11) 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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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.

(15) 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) = x1
f(x1, x2, x3) = f(x2, x3)
ACTIVE(x1) = x1
mark(x1) = x1
a = a
b = b
active(x1) = x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[f₂, a, b]

Status:

f₂: [2,1]
a: []
b: []

The following usable rules [FROCOS05] were oriented:

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)

(16) 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.