Termination w.r.t. Q proof of /tmp/tmpI37OxV/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 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 PisEmptyProof (⇔)

    ↳30 TRUE

  ↳31 QDP

    ↳32 QDPOrderProof (⇔)

    ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 PisEmptyProof (⇔)

    ↳37 TRUE

  ↳38 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(h(X)) → G(X, X)
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(f(X, X)) → H(a)
ACTIVE(h(X)) → H(active(X))
ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(g(X1, X2)) → G(active(X1), X2)
ACTIVE(g(X1, X2)) → ACTIVE(X1)
ACTIVE(f(X1, X2)) → F(active(X1), X2)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
H(mark(X)) → H(X)
G(mark(X1), X2) → G(X1, X2)
F(mark(X1), X2) → F(X1, X2)
PROPER(h(X)) → H(proper(X))
PROPER(h(X)) → PROPER(X)
PROPER(g(X1, X2)) → G(proper(X1), proper(X2))
PROPER(g(X1, X2)) → PROPER(X1)
PROPER(g(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → F(proper(X1), proper(X2))
PROPER(f(X1, X2)) → PROPER(X1)
PROPER(f(X1, X2)) → PROPER(X2)
H(ok(X)) → H(X)
G(ok(X1), ok(X2)) → G(X1, X2)
F(ok(X1), ok(X2)) → F(X1, X2)
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(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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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 6 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(ok(X1), ok(X2)) → F(X1, X2)
F(mark(X1), 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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)) → 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(x2)
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = active(x1)
h(x1) = x1
g(x1, x2) = x2
a = a
f(x1, x2) = x1
b = b
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

top > active₁ > [a, b]
top > proper₁ > ok₁ > F₁
top > proper₁ > [a, b]

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

F(mark(X1), 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) 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) = active(x1)
h(x1) = h(x1)
g(x1, x2) = x1
a = a
f(x1, x2) = f(x1)
b = b
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[active₁, a] > f₁ > h₁ > mark₁ > F₂
[active₁, a] > b > F₂
top > F₂

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(9) 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

G(ok(X1), ok(X2)) → G(X1, X2)
G(mark(X1), 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(X1), ok(X2)) → G(X1, X2)

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

Recursive Path Order [RPO].
Precedence:

top > active₁ > [a, b]
top > proper₁ > ok₁ > G₁
top > proper₁ > [a, b]

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(14) Obligation:

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

G(mark(X1), 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

G(mark(X1), X2) → G(X1, X2)

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

Recursive Path Order [RPO].
Precedence:

[active₁, a] > f₁ > h₁ > mark₁ > G₂
[active₁, a] > b > G₂
top > G₂

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(16) 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

H(ok(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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

H(ok(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
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = active(x1)
h(x1) = h(x1)
g(x1, x2) = x2
a = a
f(x1, x2) = f(x1)
b = b
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[active₁, proper₁, top] > [h₁, f₁] > ok₁ > a
[active₁, proper₁, top] > b > a

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(21) 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(22) 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: Combined order from the following AFS and order.
H(x1) = H(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
h(x1) = x1
g(x1, x2) = x1
a = a
f(x1, x2) = x1
b = b
proper(x1) = proper
ok(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[H₁, mark₁, active₁, proper, top] > [a, b]

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(23) 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Recursive Path Order [RPO].
Precedence:

[g₂, active₁, top] > [h₁, f₂] > a
[g₂, active₁, top] > b > a

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(28) 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) TRUE

(31) Obligation:

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

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(g(X1, X2)) → ACTIVE(X1)

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, x2) = g(x1)
h(x1) = x1
f(x1, x2) = x1
active(x1) = active(x1)
mark(x1) = mark
a = a
b = b
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[ACTIVE₁, g₁, active₁, mark, b, top] > a

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(33) Obligation:

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

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X1, X2)) → ACTIVE(X1)

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

Recursive Path Order [RPO].
Precedence:

[f₁, g₁] > [ACTIVE₁, h₁] > mark > top > b
[f₁, g₁] > a > mark > top > b

The following usable rules [FROCOS05] were oriented:

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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(35) 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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE

(38) 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(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)
active(h(X)) → h(active(X))
active(g(X1, X2)) → g(active(X1), X2)
active(f(X1, X2)) → f(active(X1), X2)
h(mark(X)) → mark(h(X))
g(mark(X1), X2) → mark(g(X1, X2))
f(mark(X1), X2) → mark(f(X1, X2))
proper(h(X)) → h(proper(X))
proper(g(X1, X2)) → g(proper(X1), proper(X2))
proper(a) → ok(a)
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(b) → ok(b)
h(ok(X)) → ok(h(X))
g(ok(X1), ok(X2)) → ok(g(X1, X2))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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.