(0) Obligation:

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(0)) → CONS(0, f(s(0)))
ACTIVE(f(0)) → F(s(0))
ACTIVE(f(0)) → S(0)
ACTIVE(f(s(0))) → F(p(s(0)))
ACTIVE(f(s(0))) → P(s(0))
ACTIVE(f(X)) → F(active(X))
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(p(X)) → P(active(X))
ACTIVE(p(X)) → ACTIVE(X)
F(mark(X)) → F(X)
CONS(mark(X1), X2) → CONS(X1, X2)
S(mark(X)) → S(X)
P(mark(X)) → P(X)
PROPER(f(X)) → F(proper(X))
PROPER(f(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(p(X)) → P(proper(X))
PROPER(p(X)) → PROPER(X)
F(ok(X)) → F(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
S(ok(X)) → S(X)
P(ok(X)) → P(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(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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 7 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

P(ok(X)) → P(X)
P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.


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

Recursive path order with status [RPO].
Precedence:
top > proper > ok1
top > proper > 0

Status:
P1: multiset
ok1: [1]
0: multiset
proper: []
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.


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

Recursive path order with status [RPO].
Precedence:
top > active1 > 0 > mark1
top > active1 > cons1 > ok > mark1
top > active1 > s1 > mark1
top > proper1 > 0 > mark1
top > proper1 > cons1 > ok > mark1
top > proper1 > s1 > mark1

Status:
mark1: [1]
active1: [1]
0: multiset
cons1: [1]
s1: [1]
proper1: multiset
ok: []
top: multiset

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.


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

Recursive path order with status [RPO].
Precedence:
top > proper > ok1
top > proper > 0

Status:
S1: multiset
ok1: [1]
0: multiset
proper: []
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.


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

Recursive path order with status [RPO].
Precedence:
top > active1 > 0 > mark1
top > active1 > cons1 > ok > mark1
top > active1 > s1 > mark1
top > proper1 > 0 > mark1
top > proper1 > cons1 > ok > mark1
top > proper1 > s1 > mark1

Status:
mark1: [1]
active1: [1]
0: multiset
cons1: [1]
s1: [1]
proper1: multiset
ok: []
top: multiset

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

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

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.


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)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
0  =  0
cons(x1, x2)  =  x1
s(x1)  =  s(x1)
p(x1)  =  p(x1)
proper(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > f1 > 0 > mark1
active1 > s1 > 0 > mark1
active1 > p1 > 0 > mark1

Status:
CONS2: [2,1]
mark1: multiset
active1: multiset
f1: [1]
0: multiset
s1: [1]
p1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.


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)  =  CONS(x1)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
f(x1)  =  x1
0  =  0
mark(x1)  =  x1
cons(x1, x2)  =  x2
s(x1)  =  x1
p(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
proper1 > ok1
proper1 > 0
top > active1 > 0

Status:
CONS1: [1]
ok1: multiset
active1: [1]
0: multiset
proper1: multiset
top: multiset

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

F(ok(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.


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

Recursive path order with status [RPO].
Precedence:
top > proper > ok1
top > proper > 0

Status:
F1: multiset
ok1: [1]
0: multiset
proper: []
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.


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

Recursive path order with status [RPO].
Precedence:
top > active1 > 0 > mark1
top > active1 > cons1 > ok > mark1
top > active1 > s1 > mark1
top > proper1 > 0 > mark1
top > proper1 > cons1 > ok > mark1
top > proper1 > s1 > mark1

Status:
mark1: [1]
active1: [1]
0: multiset
cons1: [1]
s1: [1]
proper1: multiset
ok: []
top: multiset

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(f(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(p(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(34) 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(X1)
PROPER(cons(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)  =  PROPER(x1)
cons(x1, x2)  =  cons(x1, x2)
f(x1)  =  x1
s(x1)  =  x1
p(x1)  =  x1
active(x1)  =  active(x1)
0  =  0
mark(x1)  =  mark
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > cons2 > mark > PROPER1
active1 > 0 > PROPER1
proper1 > cons2 > mark > PROPER1
proper1 > 0 > PROPER1
proper1 > ok > top > PROPER1

Status:
PROPER1: multiset
cons2: multiset
active1: multiset
0: multiset
mark: multiset
proper1: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(35) Obligation:

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

PROPER(f(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(p(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive path order with status [RPO].
Precedence:
0 > p1 > mark
proper1 > p1 > mark
top > mark

Status:
p1: [1]
0: multiset
mark: multiset
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(37) Obligation:

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

PROPER(f(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(f(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)  =  f(x1)
s(x1)  =  x1
active(x1)  =  x1
0  =  0
mark(x1)  =  x1
cons(x1, x2)  =  x2
p(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
PROPER1 > f1
top > proper1 > 0 > f1

Status:
PROPER1: multiset
f1: [1]
0: multiset
proper1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(39) Obligation:

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

PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(s(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)
s(x1)  =  s(x1)
active(x1)  =  active(x1)
f(x1)  =  f(x1)
0  =  0
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1)
p(x1)  =  p
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > s1 > f1
active1 > 0 > f1
active1 > cons1
active1 > p
proper1 > s1 > f1
proper1 > 0 > f1
proper1 > cons1
proper1 > p

Status:
PROPER1: [1]
s1: multiset
active1: [1]
f1: multiset
0: multiset
cons1: multiset
p: multiset
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(41) Obligation:

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(p(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(p(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)
cons(x1, x2)  =  x1
f(x1)  =  x1
s(x1)  =  x1
p(x1)  =  p(x1)
active(x1)  =  x1
0  =  0
mark(x1)  =  mark
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
ACTIVE1 > top
0 > mark > p1 > top

Status:
ACTIVE1: [1]
p1: [1]
0: multiset
mark: multiset
top: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(46) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(s(X)) → ACTIVE(X)
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)  =  x1
f(x1)  =  x1
s(x1)  =  s(x1)
active(x1)  =  x1
0  =  0
mark(x1)  =  x1
p(x1)  =  p
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top(x1)

Recursive path order with status [RPO].
Precedence:
s1 > p > 0

Status:
s1: [1]
0: multiset
p: multiset
top1: multiset

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(48) Obligation:

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

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

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(49) 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)  =  ACTIVE(x1)
cons(x1, x2)  =  cons(x1)
f(x1)  =  x1
active(x1)  =  active(x1)
0  =  0
mark(x1)  =  x1
s(x1)  =  x1
p(x1)  =  p
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
ACTIVE1 > ok
active1 > cons1 > ok
active1 > 0 > ok
active1 > p > ok
proper1 > cons1 > ok
proper1 > 0 > ok
proper1 > p > ok
top > ok

Status:
ACTIVE1: multiset
cons1: multiset
active1: [1]
0: multiset
p: multiset
proper1: [1]
ok: multiset
top: multiset

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(50) Obligation:

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

ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(f(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)
f(x1)  =  f(x1)
active(x1)  =  x1
0  =  0
mark(x1)  =  x1
cons(x1, x2)  =  cons
s(x1)  =  s(x1)
p(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  x1

Recursive path order with status [RPO].
Precedence:
f1 > 0 > cons
s1 > 0 > cons

Status:
ACTIVE1: multiset
f1: multiset
0: multiset
cons: multiset
s1: multiset

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(52) Obligation:

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) 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(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(0))) → mark(0)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.