(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(length(cons(N, L))) → S(length(L))
ACTIVE(length(cons(N, L))) → LENGTH(L)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → LENGTH(active(X))
ACTIVE(length(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
S(mark(X)) → S(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(length(X)) → LENGTH(proper(X))
PROPER(length(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
AND(ok(X1), ok(X2)) → AND(X1, X2)
LENGTH(ok(X)) → LENGTH(X)
S(ok(X)) → S(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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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 13 less nodes.

(4) Complex Obligation (AND)

(5) 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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.


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)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x2)
0  =  0
and(x1, x2)  =  x2
tt  =  tt
length(x1)  =  x1
nil  =  nil
s(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
S1 > [ok1, zeros, cons1, 0, nil, proper1]
[active1, top] > [ok1, zeros, cons1, 0, nil, proper1]
tt > [ok1, zeros, cons1, 0, nil, proper1]

Status:
S1: multiset
ok1: [1]
active1: [1]
zeros: multiset
cons1: [1]
0: multiset
tt: multiset
nil: multiset
proper1: [1]
top: []


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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:

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.


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)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
and(x1, x2)  =  and(x1, x2)
tt  =  tt
length(x1)  =  x1
nil  =  nil
s(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
zeros > [active1, 0] > cons2 > [tt, ok] > mark1
zeros > [active1, 0] > and2 > [tt, ok] > mark1
nil > [tt, ok] > mark1
top > mark1

Status:
mark1: [1]
active1: multiset
zeros: multiset
cons2: multiset
0: multiset
and2: [1,2]
tt: multiset
nil: multiset
ok: []
top: multiset


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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:

LENGTH(ok(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.


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

Recursive path order with status [RPO].
Quasi-Precedence:
LENGTH1 > [ok1, zeros, cons1, 0, nil, proper1]
[active1, top] > [ok1, zeros, cons1, 0, nil, proper1]
tt > [ok1, zeros, cons1, 0, nil, proper1]

Status:
LENGTH1: multiset
ok1: [1]
active1: [1]
zeros: multiset
cons1: [1]
0: multiset
tt: multiset
nil: multiset
proper1: [1]
top: []


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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:

LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.


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

Recursive path order with status [RPO].
Quasi-Precedence:
zeros > [active1, 0] > cons2 > [tt, ok] > mark1
zeros > [active1, 0] > and2 > [tt, ok] > mark1
nil > [tt, ok] > mark1
top > mark1

Status:
mark1: [1]
active1: multiset
zeros: multiset
cons2: multiset
0: multiset
and2: [1,2]
tt: multiset
nil: multiset
ok: []
top: multiset


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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:

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.


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

Recursive path order with status [RPO].
Quasi-Precedence:
zeros > [0, proper1] > and1 > [ok1, mark, tt]
nil > [0, proper1] > and1 > [ok1, mark, tt]
top > [ok1, mark, tt]

Status:
ok1: [1]
mark: []
zeros: multiset
0: multiset
and1: [1]
tt: multiset
nil: multiset
proper1: [1]
top: []


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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:

AND(mark(X1), X2) → AND(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.


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

Recursive path order with status [RPO].
Quasi-Precedence:
AND1 > top
active1 > zeros > top
active1 > cons2 > [mark1, tt] > top
active1 > [0, nil] > top
active1 > and2 > [mark1, tt] > top

Status:
AND1: multiset
mark1: [1]
active1: [1]
zeros: multiset
cons2: multiset
0: multiset
and2: multiset
tt: multiset
nil: multiset
top: []


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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:

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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.


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)  =  x2
ok(x1)  =  ok(x1)
mark(x1)  =  mark
active(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
and(x1, x2)  =  and(x1)
tt  =  tt
length(x1)  =  x1
nil  =  nil
s(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
zeros > [0, proper1] > and1 > [ok1, mark, tt]
nil > [0, proper1] > and1 > [ok1, mark, tt]
top > [ok1, mark, tt]

Status:
ok1: [1]
mark: []
zeros: multiset
0: multiset
and1: [1]
tt: multiset
nil: multiset
proper1: [1]
top: []


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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:

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.


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)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
and(x1, x2)  =  and(x1, x2)
tt  =  tt
length(x1)  =  x1
nil  =  nil
s(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
CONS1 > top
active1 > zeros > top
active1 > cons2 > [mark1, tt] > top
active1 > [0, nil] > top
active1 > and2 > [mark1, tt] > top

Status:
CONS1: multiset
mark1: [1]
active1: [1]
zeros: multiset
cons2: multiset
0: multiset
and2: multiset
tt: multiset
nil: multiset
top: []


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(length(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(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)
and(x1, x2)  =  and(x1, x2)
length(x1)  =  x1
s(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark
0  =  0
tt  =  tt
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
active1 > and2 > ok > [PROPER1, cons2] > [zeros, mark, 0]
[proper1, top] > and2 > ok > [PROPER1, cons2] > [zeros, mark, 0]
[proper1, top] > tt > [zeros, mark, 0]
[proper1, top] > nil > [zeros, mark, 0]

Status:
PROPER1: [1]
cons2: multiset
and2: multiset
active1: [1]
zeros: multiset
mark: multiset
0: multiset
tt: multiset
nil: multiset
proper1: [1]
ok: []
top: multiset


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(length(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(length(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)
length(x1)  =  length(x1)
s(x1)  =  x1
active(x1)  =  x1
zeros  =  zeros
mark(x1)  =  mark
cons(x1, x2)  =  x1
0  =  0
and(x1, x2)  =  and
tt  =  tt
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
[and, proper1] > zeros > mark > top
[and, proper1] > zeros > 0
[and, proper1] > tt > mark > top
[and, proper1] > nil > mark > top
[and, proper1] > nil > 0
[and, proper1] > ok > [PROPER1, length1] > mark > top
[and, proper1] > ok > [PROPER1, length1] > 0

Status:
PROPER1: multiset
length1: multiset
zeros: multiset
mark: multiset
0: multiset
and: multiset
tt: multiset
nil: multiset
proper1: [1]
ok: []
top: []


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(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)  =  x1
zeros  =  zeros
mark(x1)  =  mark
cons(x1, x2)  =  cons(x2)
0  =  0
and(x1, x2)  =  and
tt  =  tt
length(x1)  =  length(x1)
nil  =  nil
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
PROPER1 > [s1, mark, and]
zeros > [cons1, length1] > 0 > [s1, mark, and]
tt > [s1, mark, and]
nil > [s1, mark, and]
top > [s1, mark, and]

Status:
PROPER1: multiset
s1: [1]
zeros: multiset
mark: multiset
cons1: [1]
0: multiset
and: multiset
tt: multiset
length1: [1]
nil: multiset
top: multiset


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(39) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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) PisEmptyProof (EQUIVALENT transformation)

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

(41) TRUE

(42) Obligation:

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

ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)
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
and(x1, x2)  =  and(x1)
cons(x1, x2)  =  cons(x1)
length(x1)  =  length(x1)
s(x1)  =  s(x1)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark
0  =  0
tt  =  tt
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Quasi-Precedence:
top > [and1, cons1, s1, active1, proper1] > length1 > [0, nil] > [zeros, mark]
top > [and1, cons1, s1, active1, proper1] > length1 > ok1
top > [and1, cons1, s1, active1, proper1] > tt > [zeros, mark]

Status:
and1: [1]
cons1: [1]
length1: [1]
s1: [1]
active1: [1]
zeros: multiset
mark: multiset
0: multiset
tt: multiset
nil: multiset
proper1: [1]
ok1: multiset
top: multiset


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(44) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) 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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.