(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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(zeros) → CONS(0, zeros)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → TAIL(active(X))
ACTIVE(tail(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
TAIL(mark(X)) → TAIL(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(tail(X)) → TAIL(proper(X))
PROPER(tail(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
TAIL(ok(X)) → TAIL(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(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

TAIL(ok(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.


TAIL(ok(X)) → TAIL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAIL(x1)  =  TAIL(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
tail(x1)  =  x1
proper(x1)  =  proper
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
TAIL1 > active1
top > proper > zeros > ok1 > active1
top > proper > 0 > ok1 > active1

Status:
active1: [1]
zeros: []
proper: []
ok1: [1]
top: []
TAIL1: [1]
0: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
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:

TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TAIL(mark(X)) → TAIL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAIL(x1)  =  TAIL(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
tail(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
TAIL1 > top
active1 > zeros > cons2 > mark1 > top
active1 > zeros > cons2 > ok > top
active1 > 0 > ok > top
proper1 > zeros > cons2 > mark1 > top
proper1 > zeros > cons2 > ok > top
proper1 > 0 > ok > top

Status:
active1: [1]
cons2: [2,1]
zeros: []
mark1: [1]
ok: []
proper1: [1]
TAIL1: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
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(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.

(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:

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.


CONS(ok(X1), ok(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
tail(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
top > active1 > zeros
top > proper1 > ok1
top > proper1 > zeros
top > proper1 > 0

Status:
active1: [1]
zeros: []
ok1: [1]
proper1: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
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:

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(mark(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  CONS(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
tail(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > cons2 > mark1 > CONS2
active1 > 0 > CONS2
zeros > cons2 > mark1 > CONS2
zeros > 0 > CONS2
top > CONS2

Status:
cons2: [1,2]
active1: [1]
zeros: []
CONS2: [2,1]
mark1: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
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(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.

(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:

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(tail(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
cons(x1, x2)  =  cons(x1, x2)
tail(x1)  =  x1
active(x1)  =  x1
zeros  =  zeros
mark(x1)  =  mark
0  =  0
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
PROPER1 > ok
proper1 > zeros > cons2 > mark > ok
proper1 > zeros > 0 > ok
top > ok

Status:
cons2: [1,2]
PROPER1: [1]
zeros: []
mark: []
ok: []
proper1: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
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:

PROPER(tail(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(tail(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)
tail(x1)  =  tail(x1)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons(x1, x2)
0  =  0
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
PROPER1 > top
active1 > tail1 > mark1 > top
active1 > cons2 > mark1 > top
zeros > 0 > ok > cons2 > mark1 > top

Status:
PROPER1: [1]
active1: [1]
cons2: [1,2]
tail1: [1]
zeros: []
mark1: [1]
ok: []
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
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(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.

(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:

ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(tail(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)
tail(x1)  =  tail(x1)
cons(x1, x2)  =  x1
active(x1)  =  x1
zeros  =  zeros
mark(x1)  =  mark
0  =  0
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
0 > ok > tail1 > mark
proper1 > zeros > mark
proper1 > ok > tail1 > mark

Status:
tail1: [1]
zeros: []
mark: []
ok: []
proper1: [1]
top: []
0: []
ACTIVE1: [1]

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(28) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(cons(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)  =  x1
cons(x1, x2)  =  cons(x1, x2)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
0  =  0
tail(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > 0 > ok > cons2
zeros > 0 > ok > cons2
top > proper1 > cons2

Status:
active1: [1]
cons2: [2,1]
zeros: []
ok: []
proper1: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(30) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) 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(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(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.