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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 PisEmptyProof (⇔)
    ↳25 TRUE
  ↳26 QDP
    ↳27 QDPOrderProof (⇔)
    ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 PisEmptyProof (⇔)
    ↳32 TRUE
  ↳33 QDP
    ↳34 QDPOrderProof (⇔)
    ↳35 QDP
    ↳36 QDPOrderProof (⇔)
    ↳37 QDP
    ↳38 PisEmptyProof (⇔)
    ↳39 TRUE
  ↳40 QDP
    ↳41 QDPOrderProof (⇔)
    ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 PisEmptyProof (⇔)
    ↳48 TRUE
  ↳49 QDP
    ↳50 QDPOrderProof (⇔)
    ↳51 QDP
    ↳52 QDPOrderProof (⇔)
    ↳53 QDP
    ↳54 QDPOrderProof (⇔)
    ↳55 QDP
    ↳56 QDPOrderProof (⇔)
    ↳57 QDP
    ↳58 QDPOrderProof (⇔)
    ↳59 QDP
    ↳60 PisEmptyProof (⇔)
    ↳61 TRUE
  ↳62 QDP

(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(U11(tt, L)) → U121(tt, L)
ACTIVE(U12(tt, L)) → S(length(L))
ACTIVE(U12(tt, L)) → LENGTH(L)
ACTIVE(length(cons(N, L))) → U111(tt, L)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(U11(X1, X2)) → U111(active(X1), X2)
ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(U12(X1, X2)) → U121(active(X1), X2)
ACTIVE(U12(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(length(X)) → LENGTH(active(X))
ACTIVE(length(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
U111(mark(X1), X2) → U111(X1, X2)
U121(mark(X1), X2) → U121(X1, X2)
S(mark(X)) → S(X)
LENGTH(mark(X)) → LENGTH(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(U11(X1, X2)) → U111(proper(X1), proper(X2))
PROPER(U11(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U12(X1, X2)) → U121(proper(X1), proper(X2))
PROPER(U12(X1, X2)) → PROPER(X1)
PROPER(U12(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → LENGTH(proper(X))
PROPER(length(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
U111(ok(X1), ok(X2)) → U111(X1, X2)
U121(ok(X1), ok(X2)) → U121(X1, X2)
S(ok(X)) → S(X)
LENGTH(ok(X)) → LENGTH(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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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 8 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) 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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


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)  =  LENGTH(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  x1
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
zeros > cons2 > mark1 > LENGTH1
zeros > 0 > LENGTH1
top > active1 > cons2 > mark1 > LENGTH1
top > active1 > 0 > LENGTH1
top > active1 > U112 > mark1 > LENGTH1
top > active1 > tt > U122 > mark1 > LENGTH1
top > proper1 > cons2 > mark1 > LENGTH1
top > proper1 > 0 > LENGTH1
top > proper1 > U112 > mark1 > LENGTH1
top > proper1 > tt > U122 > mark1 > LENGTH1
top > proper1 > nil > LENGTH1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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:

LENGTH(ok(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


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)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  x2
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x2)
s(x1)  =  x1
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros
active1 > 0 > ok1
active1 > U112 > ok1
active1 > tt > U121 > ok1
top > proper1 > zeros
top > proper1 > 0 > ok1
top > proper1 > U112 > ok1
top > proper1 > U121 > ok1
top > proper1 > nil

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(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)  =  S(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  x1
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
zeros > cons2 > mark1 > S1
zeros > 0 > S1
top > active1 > cons2 > mark1 > S1
top > active1 > 0 > S1
top > active1 > U112 > mark1 > S1
top > active1 > tt > U122 > mark1 > S1
top > proper1 > cons2 > mark1 > S1
top > proper1 > 0 > S1
top > proper1 > U112 > mark1 > S1
top > proper1 > tt > U122 > mark1 > S1
top > proper1 > nil > S1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(ok(X)) → S(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(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)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  x2
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x2)
s(x1)  =  x1
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros
active1 > 0 > ok1
active1 > U112 > ok1
active1 > tt > U121 > ok1
top > proper1 > zeros
top > proper1 > 0 > ok1
top > proper1 > U112 > ok1
top > proper1 > U121 > ok1
top > proper1 > nil

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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:

U121(ok(X1), ok(X2)) → U121(X1, X2)
U121(mark(X1), X2) → U121(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


U121(ok(X1), ok(X2)) → U121(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U121(x1, x2)  =  U121(x2)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  s(x1)
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
U12^11 > ok1
proper1 > zeros > 0 > ok1
proper1 > cons2 > U112 > ok1
proper1 > tt > ok1
proper1 > U122 > ok1
proper1 > s1 > ok1
proper1 > nil > ok1
top > active1 > zeros > 0 > ok1
top > active1 > cons2 > U112 > ok1
top > active1 > tt > ok1
top > active1 > U122 > ok1
top > active1 > s1 > ok1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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:

U121(mark(X1), X2) → U121(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


U121(mark(X1), X2) → U121(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U121(x1, x2)  =  U121(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  x1
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > cons2 > U112 > mark1 > U12^11
active1 > cons2 > U112 > tt > U12^11
active1 > 0 > U12^11
active1 > U122 > mark1 > U12^11
zeros > cons2 > U112 > mark1 > U12^11
zeros > cons2 > U112 > tt > U12^11
zeros > 0 > U12^11
proper1 > cons2 > U112 > mark1 > U12^11
proper1 > cons2 > U112 > tt > U12^11
proper1 > 0 > U12^11
proper1 > U122 > mark1 > U12^11
proper1 > nil > U12^11
top > U12^11

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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:

U111(ok(X1), ok(X2)) → U111(X1, X2)
U111(mark(X1), X2) → U111(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


U111(ok(X1), ok(X2)) → U111(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U111(x1, x2)  =  U111(x2)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  s(x1)
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
U11^11 > ok1
proper1 > zeros > 0 > ok1
proper1 > cons2 > U112 > ok1
proper1 > tt > ok1
proper1 > U122 > ok1
proper1 > s1 > ok1
proper1 > nil > ok1
top > active1 > zeros > 0 > ok1
top > active1 > cons2 > U112 > ok1
top > active1 > tt > ok1
top > active1 > U122 > ok1
top > active1 > s1 > ok1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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:

U111(mark(X1), X2) → U111(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


U111(mark(X1), X2) → U111(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U111(x1, x2)  =  U111(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  x1
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > cons2 > U112 > mark1 > U11^11
active1 > cons2 > U112 > tt > U11^11
active1 > 0 > U11^11
active1 > U122 > mark1 > U11^11
zeros > cons2 > U112 > mark1 > U11^11
zeros > cons2 > U112 > tt > U11^11
zeros > 0 > U11^11
proper1 > cons2 > U112 > mark1 > U11^11
proper1 > cons2 > U112 > tt > U11^11
proper1 > 0 > U11^11
proper1 > U122 > mark1 > U11^11
proper1 > nil > U11^11
top > U11^11

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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:

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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


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(x2)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  s(x1)
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
CONS1 > ok1
proper1 > zeros > 0 > ok1
proper1 > cons2 > U112 > ok1
proper1 > tt > ok1
proper1 > U122 > ok1
proper1 > s1 > ok1
proper1 > nil > ok1
top > active1 > zeros > 0 > ok1
top > active1 > cons2 > U112 > ok1
top > active1 > tt > ok1
top > active1 > U122 > ok1
top > active1 > s1 > ok1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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:

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


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
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  x1
length(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > cons2 > U112 > mark1 > CONS1
active1 > cons2 > U112 > tt > CONS1
active1 > 0 > CONS1
active1 > U122 > mark1 > CONS1
zeros > cons2 > U112 > mark1 > CONS1
zeros > cons2 > U112 > tt > CONS1
zeros > 0 > CONS1
proper1 > cons2 > U112 > mark1 > CONS1
proper1 > cons2 > U112 > tt > CONS1
proper1 > 0 > CONS1
proper1 > U122 > mark1 > CONS1
proper1 > nil > CONS1
top > CONS1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(37) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) 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(U11(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U12(X1, X2)) → PROPER(X1)
PROPER(U12(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.

(41) 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(U11(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U12(X1, X2)) → PROPER(X1)
PROPER(U12(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)
U11(x1, x2)  =  U11(x1, x2)
U12(x1, x2)  =  U12(x1, x2)
s(x1)  =  x1
length(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark(x1)
0  =  0
tt  =  tt
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > cons2 > mark1
active1 > U112 > tt > mark1
active1 > U112 > tt > ok
active1 > U122 > mark1
active1 > U122 > ok
active1 > 0
zeros > cons2 > mark1
zeros > 0
zeros > ok
proper1 > cons2 > mark1
proper1 > U112 > tt > mark1
proper1 > U112 > tt > ok
proper1 > U122 > mark1
proper1 > U122 > ok
proper1 > nil > mark1
proper1 > nil > 0

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(42) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


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)  =  x1
s(x1)  =  s(x1)
length(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  x2
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U12(x1, x2)  =  x2
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
proper1 > s1 > U112
proper1 > zeros > U112
proper1 > 0 > U112
proper1 > tt > U112
proper1 > nil > U112
top > active1 > s1 > U112
top > active1 > zeros > U112
top > active1 > 0 > U112
top > active1 > tt > U112

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(44) Obligation:

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

PROPER(length(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.


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)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  cons
0  =  0
U11(x1, x2)  =  x1
tt  =  tt
U12(x1, x2)  =  x1
s(x1)  =  s
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
PROPER1 > length1
zeros > cons > tt > s > active1 > 0 > length1
proper1 > cons > tt > s > active1 > 0 > length1
proper1 > nil > length1
top > length1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(46) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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) PisEmptyProof (EQUIVALENT transformation)

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

(48) TRUE

(49) Obligation:

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

ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(U12(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(length(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.

(50) 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)  =  ACTIVE(x1)
U11(x1, x2)  =  x1
cons(x1, x2)  =  x1
U12(x1, x2)  =  x1
s(x1)  =  s(x1)
length(x1)  =  x1
active(x1)  =  x1
zeros  =  zeros
mark(x1)  =  mark
0  =  0
tt  =  tt
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
zeros > mark
0 > mark
nil > mark
top > proper1 > s1 > ACTIVE1 > mark
top > proper1 > tt > mark

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(51) Obligation:

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

ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(U12(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.

(52) 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)
U11(x1, x2)  =  x1
cons(x1, x2)  =  cons(x1, x2)
U12(x1, x2)  =  x1
length(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark
0  =  0
tt  =  tt
s(x1)  =  s(x1)
nil  =  nil
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > cons2 > mark > ACTIVE1
active1 > cons2 > ok > ACTIVE1
active1 > zeros > ok > ACTIVE1
active1 > s1 > mark > ACTIVE1
active1 > s1 > ok > ACTIVE1
0 > ok > ACTIVE1
tt > s1 > mark > ACTIVE1
tt > s1 > ok > ACTIVE1
nil > mark > ACTIVE1
nil > ok > ACTIVE1
top > ACTIVE1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(53) Obligation:

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

ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(U12(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(U11(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)
U11(x1, x2)  =  U11(x1, x2)
U12(x1, x2)  =  x1
length(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
0  =  0
tt  =  tt
s(x1)  =  s
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > cons2 > U112 > ok
active1 > 0 > ok
active1 > tt > s > ok
zeros > cons2 > U112 > ok
zeros > 0 > ok
top > proper1 > cons2 > U112 > ok
top > proper1 > 0 > ok
top > proper1 > s > ok
top > proper1 > nil

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(55) Obligation:

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

ACTIVE(U12(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(length(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)
U12(x1, x2)  =  x1
length(x1)  =  length(x1)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
s(x1)  =  s(x1)
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros > ok1
active1 > mark > length1 > ok1
active1 > mark > cons2 > U112 > ok1
active1 > mark > top
active1 > s1 > ok1
0 > ok1
tt > mark > length1 > ok1
tt > mark > cons2 > U112 > ok1
tt > mark > top
tt > s1 > ok1
proper1 > length1 > ok1
proper1 > zeros > ok1
proper1 > cons2 > U112 > ok1
proper1 > s1 > ok1
proper1 > nil > ok1

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(57) Obligation:

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

ACTIVE(U12(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(U12(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)
U12(x1, x2)  =  U12(x1, x2)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
s(x1)  =  x1
length(x1)  =  length(x1)
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros
active1 > cons2 > U112
active1 > 0
active1 > tt > U122 > ACTIVE1
active1 > tt > U122 > length1 > U112
proper1 > U122 > ACTIVE1
proper1 > U122 > length1 > U112
proper1 > zeros
proper1 > cons2 > U112
proper1 > 0
proper1 > nil

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(59) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.

(60) PisEmptyProof (EQUIVALENT transformation)

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

(61) TRUE

(62) 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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(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.