Termination w.r.t. Q proof of /tmp/tmpZwNVOc/Ex14_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 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

(0) Obligation:

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

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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(h(X)) → MARK(g(X, X))
ACTIVE(h(X)) → G(X, X)
ACTIVE(g(a, X)) → MARK(f(b, X))
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(f(X, X)) → MARK(h(a))
ACTIVE(f(X, X)) → H(a)
ACTIVE(a) → MARK(b)
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → H(mark(X))
MARK(h(X)) → MARK(X)
MARK(g(X1, X2)) → ACTIVE(g(mark(X1), X2))
MARK(g(X1, X2)) → G(mark(X1), X2)
MARK(g(X1, X2)) → MARK(X1)
MARK(a) → ACTIVE(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(b) → ACTIVE(b)
H(mark(X)) → H(X)
H(active(X)) → H(X)
G(mark(X1), X2) → G(X1, X2)
G(X1, mark(X2)) → G(X1, X2)
G(active(X1), X2) → G(X1, X2)
G(X1, active(X2)) → G(X1, X2)
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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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 4 SCCs with 9 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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)
F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > F₂
active₁ > F₂

Status:

F₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > G₂
active₁ > G₂

Status:

G₂: [1,2]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

H(active(X)) → H(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
H(x1) = x1
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

H(mark(X)) → H(X)

The TRS R consists of the following rules:

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

H(mark(X)) → H(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > H₁

Status:

H₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

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

The TRS R consists of the following rules:

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