(0) Obligation:

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

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

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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(X1, mark(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(x2)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1, x2, x3)  =  x2
b  =  b
c  =  c
proper(x1)  =  proper
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
F1 > mark1
active1 > c > b > mark1
proper > c > b > mark1
top > mark1

Status:
F1: [1]
mark1: [1]
active1: [1]
b: []
c: []
proper: []
top: []

The following usable rules [FROCOS05] were oriented:

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

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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(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(x2)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
f(x1, x2, x3)  =  x2
b  =  b
c  =  c
mark(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
top > active1 > c > b
top > proper1 > ok1
top > proper1 > c > b

Status:
F1: [1]
ok1: [1]
active1: [1]
b: []
c: []
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
f(x1, x2, x3)  =  f(x1, x2, x3)
active(x1)  =  active(x1)
b  =  b
c  =  c
mark(x1)  =  mark
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > mark > f3 > c
active1 > mark > f3 > ok1
b > mark > f3 > c
b > mark > f3 > ok1
top > proper1 > f3 > c
top > proper1 > f3 > ok1

Status:
f3: [2,3,1]
active1: [1]
b: []
c: []
mark: []
proper1: [1]
ok1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

The TRS R consists of the following rules:

active(f(b, X, c)) → mark(f(X, c, X))
active(c) → mark(b)
active(f(X1, X2, X3)) → f(X1, active(X2), X3)
f(X1, mark(X2), X3) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
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(X1, X2, X3)) → ACTIVE(X2)
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)
active(x1)  =  active(x1)
b  =  b
c  =  c
mark(x1)  =  mark
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
c > b > f2
c > mark > f2
top > active1 > b > f2
top > active1 > mark > f2
top > proper1 > b > f2

Status:
ACTIVE1: [1]
f2: [2,1]
active1: [1]
b: []
c: []
mark: []
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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