Termination w.r.t. Q proof of /tmp/tmpKeVt2C/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 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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)) → F(Y, Y, Y)
ACTIVE(g(X)) → G(active(X))
ACTIVE(g(X)) → ACTIVE(X)
G(mark(X)) → G(X)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
G(ok(X)) → G(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 5 SCCs with 6 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(ok(X1), ok(X2), ok(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

F(ok(X1), ok(X2), ok(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)
ok(x1) = ok(x1)
active(x1) = active
f(x1, x2, x3) = f(x1)
g(x1) = x1
mark(x1) = mark
b = b
c = c
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active > f₁ > ok₁ > top
active > mark > top
c > ok₁ > top
proper₁ > f₁ > ok₁ > top
proper₁ > b

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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

G(ok(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

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) = G(x1)
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1, x2, x3) = f(x1, x3)
g(x1) = x1
b = b
c = c
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

G₁ > mark₁
active₁ > f₂ > mark₁
b > c > mark₁
proper₁ > f₂ > mark₁
top > mark₁

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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(12) Obligation:

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

G(ok(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

G(ok(X)) → G(X)

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

Recursive Path Order [RPO].
Precedence:

G₁ > mark
top > proper > ok₁ > mark
top > proper > b > c > mark

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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(g(X)) → PROPER(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(g(X)) → PROPER(X)

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

Recursive Path Order [RPO].
Precedence:

c > g₁
top > proper₁ > f₃ > PROPER₁ > g₁
top > proper₁ > f₃ > mark > g₁
top > proper₁ > b > mark > g₁

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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(19) 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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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:

ACTIVE(g(X)) → ACTIVE(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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(g(X)) → ACTIVE(X)

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

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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(24) 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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(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)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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