Termination w.r.t. Q proof of /tmp/tmpkSR1SQ/Ex24_GM04

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 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

(0) Obligation:

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

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
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(X, g(X), Y)) → MARK(f(Y, Y, Y))
ACTIVE(f(X, g(X), Y)) → F(Y, Y, Y)
ACTIVE(g(b)) → MARK(c)
ACTIVE(b) → MARK(c)
MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, X3))
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → G(mark(X))
MARK(g(X)) → MARK(X)
MARK(b) → ACTIVE(b)
MARK(c) → ACTIVE(c)
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)
G(mark(X)) → G(X)
G(active(X)) → G(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
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 6 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(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
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)
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
active(x1) = active(x1)
mark(x1) = mark(x1)
f(x1, x2, x3) = f
g(x1) = g
b = b
c = c

Recursive Path Order [RPO].
Precedence:

b > [active₁, mark₁, f, g] > c

The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

(7) Obligation:

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

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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:

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(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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.

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 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)
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1, x2, x3) = x3
g(x1) = g
b = b
c = c

Recursive Path Order [RPO].
Precedence:

[F₂, mark₁, active₁] > g > c
[F₂, mark₁, active₁] > b > c

The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

(12) Obligation:

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

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

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
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, 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(x1, x3)
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1, x2, x3) = f
g(x1) = g
b = b
c = c

Recursive Path Order [RPO].
Precedence:

F₂ > [g, c]
b > [mark₁, active₁, f] > [g, c]

The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

(14) Obligation:

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

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, X3))
ACTIVE(f(X, g(X), Y)) → MARK(f(Y, Y, Y))
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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.

MARK(g(X)) → ACTIVE(g(mark(X)))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK
f(x1, x2, x3) = f
ACTIVE(x1) = x1
g(x1) = g
mark(x1) = mark
active(x1) = active
b = b
c = c

Recursive Path Order [RPO].
Precedence:

[MARK, f, mark, active] > g
[MARK, f, mark, active] > b
[MARK, f, mark, active] > c

The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

(19) Obligation:

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, X3))
ACTIVE(f(X, g(X), Y)) → MARK(f(Y, Y, Y))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x1
f(x1, x2, x3) = f(x3)
ACTIVE(x1) = x1
g(x1) = g(x1)
active(x1) = x1
mark(x1) = x1
b = b
c = c

Recursive Path Order [RPO].
Precedence:

[b, c]

The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

(21) Obligation:

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

MARK(f(X1, X2, X3)) → ACTIVE(f(X1, X2, X3))
ACTIVE(f(X, g(X), Y)) → MARK(f(Y, Y, Y))

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
mark(f(X1, X2, X3)) → active(f(X1, X2, X3))
mark(g(X)) → active(g(mark(X)))
mark(b) → active(b)
mark(c) → active(c)
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)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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