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

(0) Obligation:

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

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(b, c, x)) → F(x, x, x)
ACTIVE(f(x, y, z)) → F(x, y, active(z))
ACTIVE(f(x, y, z)) → ACTIVE(z)
F(x, y, mark(z)) → F(x, y, z)
PROPER(f(x, y, z)) → F(proper(x), proper(y), proper(z))
PROPER(f(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(y)
PROPER(f(x, y, z)) → PROPER(z)
F(ok(x), ok(y), ok(z)) → F(x, y, z)
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(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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 4 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(ok(x), ok(y), ok(z)) → F(x, y, z)
F(x, y, mark(z)) → F(x, y, z)

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(x, y, mark(z)) → F(x, y, z)

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, x3)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1, x2, x3)  =  f(x1, x2, x3)
b  =  b
c  =  c
d  =  d
m(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:

F₃ > mark₁
active₁ > f₃ > mark₁
active₁ > b > mark₁
active₁ > c > mark₁
proper₁ > f₃ > mark₁
proper₁ > b > mark₁
proper₁ > c > mark₁
proper₁ > d > mark₁
top > mark₁

Status:

active₁: [1]
c: []
f₃: [2,1,3]
mark₁: [1]
b: []
d: []
proper₁: [1]
top: []
F₃: [1,3,2]

The following usable rules [FROCOS05] were oriented:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(ok(x), ok(y), ok(z)) → F(x, y, z)

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(ok(x), ok(y), ok(z)) → F(x, y, z)

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)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
f(x1, x2, x3)  =  x2
b  =  b
c  =  c
mark(x1)  =  mark
d  =  d
m(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:

active₁ > b > mark
active₁ > c > mark
proper₁ > ok₁ > F₂ > mark
proper₁ > ok₁ > top > mark
proper₁ > b > mark
proper₁ > c > mark
proper₁ > d > mark

Status:

active₁: [1]
c: []
mark: []
ok₁: [1]
b: []
d: []
proper₁: [1]
F₂: [1,2]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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:

PROPER(f(x, y, z)) → PROPER(y)
PROPER(f(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(z)

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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.

PROPER(f(x, y, z)) → PROPER(y)
PROPER(f(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(z)

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)
active(x1)  =  active(x1)
b  =  b
c  =  c
mark(x1)  =  x1
d  =  d
m(x1)  =  m(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:

active₁ > b > f₃
active₁ > c
active₁ > m₁
top > proper₁ > b > f₃
top > proper₁ > c
top > proper₁ > d

Status:

PROPER₁: [1]
active₁: [1]
c: []
m₁: [1]
f₃: [2,3,1]
b: []
d: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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:

ACTIVE(f(x, y, z)) → ACTIVE(z)

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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.

ACTIVE(f(x, y, z)) → ACTIVE(z)

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

Lexicographic path order with status [LPO].
Precedence:

d > ok > f₃
top > active₁ > b > ok > f₃
top > active₁ > c > ok > f₃
top > active₁ > m₁
top > proper₁ > f₃

Status:

active₁: [1]
c: []
m₁: [1]
f₃: [1,3,2]
b: []
d: []
ok: []
proper₁: [1]
top: []
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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:

TOP(ok(x)) → TOP(active(x))
TOP(mark(x)) → TOP(proper(x))

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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.