(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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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)) → S(length(L))
ACTIVE(U11(tt, L)) → LENGTH(L)
ACTIVE(isNat(length(V1))) → ISNATLIST(V1)
ACTIVE(isNat(s(V1))) → ISNAT(V1)
ACTIVE(isNatIList(V)) → ISNATLIST(V)
ACTIVE(isNatIList(cons(V1, V2))) → AND(isNat(V1), isNatIList(V2))
ACTIVE(isNatIList(cons(V1, V2))) → ISNAT(V1)
ACTIVE(isNatIList(cons(V1, V2))) → ISNATILIST(V2)
ACTIVE(isNatList(cons(V1, V2))) → AND(isNat(V1), isNatList(V2))
ACTIVE(isNatList(cons(V1, V2))) → ISNAT(V1)
ACTIVE(isNatList(cons(V1, V2))) → ISNATLIST(V2)
ACTIVE(length(cons(N, L))) → U111(and(isNatList(L), isNat(N)), L)
ACTIVE(length(cons(N, L))) → AND(isNatList(L), isNat(N))
ACTIVE(length(cons(N, L))) → ISNATLIST(L)
ACTIVE(length(cons(N, L))) → ISNAT(N)
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(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(length(X)) → LENGTH(active(X))
ACTIVE(length(X)) → ACTIVE(X)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
CONS(mark(X1), X2) → CONS(X1, X2)
U111(mark(X1), X2) → U111(X1, X2)
S(mark(X)) → S(X)
LENGTH(mark(X)) → LENGTH(X)
AND(mark(X1), X2) → AND(X1, X2)
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(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → LENGTH(proper(X))
PROPER(length(X)) → PROPER(X)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(isNat(X)) → ISNAT(proper(X))
PROPER(isNat(X)) → PROPER(X)
PROPER(isNatList(X)) → ISNATLIST(proper(X))
PROPER(isNatList(X)) → PROPER(X)
PROPER(isNatIList(X)) → ISNATILIST(proper(X))
PROPER(isNatIList(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
U111(ok(X1), ok(X2)) → U111(X1, X2)
S(ok(X)) → S(X)
LENGTH(ok(X)) → LENGTH(X)
AND(ok(X1), ok(X2)) → AND(X1, X2)
ISNAT(ok(X)) → ISNAT(X)
ISNATLIST(ok(X)) → ISNATLIST(X)
ISNATILIST(ok(X)) → ISNATILIST(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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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 11 SCCs with 31 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ISNATILIST(ok(X)) → ISNATILIST(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.


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

Recursive Path Order [RPO].
Precedence:
[0, U112, proper1] > zeros > [ISNATILIST1, ok1, tt, nil] > mark
[0, U112, proper1] > s1 > [ISNATILIST1, ok1, tt, nil] > mark
[0, U112, proper1] > isNat1 > [ISNATILIST1, ok1, tt, nil] > mark
top > mark


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

ISNATLIST(ok(X)) → ISNATLIST(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
[0, U112, proper1] > zeros > [ISNATLIST1, ok1, tt, nil] > mark
[0, U112, proper1] > s1 > [ISNATLIST1, ok1, tt, nil] > mark
[0, U112, proper1] > isNat1 > [ISNATLIST1, ok1, tt, nil] > mark
top > mark


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(12) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

ISNAT(ok(X)) → ISNAT(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
[0, U112, proper1] > zeros > [ISNAT1, ok1, tt, nil] > mark
[0, U112, proper1] > s1 > [ISNAT1, ok1, tt, nil] > mark
[0, U112, proper1] > isNat1 > [ISNAT1, ok1, tt, nil] > mark
top > mark


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(17) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(21) 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, x2)
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
s(x1)  =  s(x1)
length(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  x1
isNatList(x1)  =  isNatList(x1)
isNatIList(x1)  =  isNatIList(x1)
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
AND2 > isNatList1
[active1, tt, isNatIList1, proper1] > [zeros, cons2, 0] > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > [U112, s1] > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > and2 > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > nil > mark1 > isNatList1
top > isNatList1


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(22) Obligation:

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

AND(ok(X1), ok(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(23) 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)  =  AND(x1, x2)
ok(x1)  =  ok(x1)
active(x1)  =  active
zeros  =  zeros
mark(x1)  =  mark
cons(x1, x2)  =  x1
0  =  0
U11(x1, x2)  =  x1
tt  =  tt
s(x1)  =  x1
length(x1)  =  x1
and(x1, x2)  =  x1
isNat(x1)  =  x1
isNatList(x1)  =  x1
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
top > proper > zeros > [ok1, tt] > active > mark
top > proper > nil > 0 > [ok1, tt] > active > mark


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(24) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(28) 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)
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
s(x1)  =  x1
length(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  x1
isNatList(x1)  =  isNatList(x1)
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros > ok1
active1 > cons2 > and2 > ok1
active1 > [0, nil] > tt > ok1
active1 > U112 > ok1
active1 > isNatList1 > tt > ok1
active1 > isNatList1 > and2 > ok1
proper1 > zeros > ok1
proper1 > cons2 > and2 > ok1
proper1 > U112 > ok1
proper1 > isNatList1 > tt > ok1
proper1 > isNatList1 > and2 > ok1
top > ok1


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(29) 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(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(30) 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)
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
s(x1)  =  s(x1)
length(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  isNat
isNatList(x1)  =  isNatList
isNatIList(x1)  =  isNatIList
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
proper1 > [active1, tt, s1, isNat, top] > [zeros, 0] > mark1
proper1 > [active1, tt, s1, isNat, top] > isNatIList > [cons2, isNatList] > U112 > mark1
proper1 > [active1, tt, s1, isNat, top] > isNatIList > and2 > mark1
proper1 > nil


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(31) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(35) 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)
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
s(x1)  =  x1
length(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  x1
isNatList(x1)  =  isNatList(x1)
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros > ok1
active1 > cons2 > and2 > ok1
active1 > [0, nil] > tt > ok1
active1 > U112 > ok1
active1 > isNatList1 > tt > ok1
active1 > isNatList1 > and2 > ok1
proper1 > zeros > ok1
proper1 > cons2 > and2 > ok1
proper1 > U112 > ok1
proper1 > isNatList1 > tt > ok1
proper1 > isNatList1 > and2 > ok1
top > ok1


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(36) 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(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(37) 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)
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
s(x1)  =  s(x1)
length(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  isNat
isNatList(x1)  =  isNatList
isNatIList(x1)  =  isNatIList
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
proper1 > [active1, tt, s1, isNat, top] > [zeros, 0] > mark1
proper1 > [active1, tt, s1, isNat, top] > isNatIList > [cons2, isNatList] > U112 > mark1
proper1 > [active1, tt, s1, isNat, top] > isNatIList > and2 > mark1
proper1 > nil


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(38) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(42) 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, x2)
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
s(x1)  =  s(x1)
length(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  x1
isNatList(x1)  =  isNatList(x1)
isNatIList(x1)  =  isNatIList(x1)
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
U11^12 > isNatList1
[active1, tt, isNatIList1, proper1] > [zeros, cons2, 0] > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > [U112, s1] > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > and2 > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > nil > mark1 > isNatList1
top > isNatList1


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(43) Obligation:

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

U111(ok(X1), ok(X2)) → U111(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(44) 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(x1, x2)
ok(x1)  =  ok(x1)
active(x1)  =  active
zeros  =  zeros
mark(x1)  =  mark
cons(x1, x2)  =  x1
0  =  0
U11(x1, x2)  =  x1
tt  =  tt
s(x1)  =  x1
length(x1)  =  x1
and(x1, x2)  =  x1
isNat(x1)  =  x1
isNatList(x1)  =  x1
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
top > proper > zeros > [ok1, tt] > active > mark
top > proper > nil > 0 > [ok1, tt] > active > mark


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(45) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
CONS2 > isNatList1
[active1, tt, isNatIList1, proper1] > [zeros, cons2, 0] > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > [U112, s1] > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > and2 > mark1 > isNatList1
[active1, tt, isNatIList1, proper1] > nil > mark1 > isNatList1
top > isNatList1


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(50) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
top > proper > zeros > [ok1, tt] > active > mark
top > proper > nil > 0 > [ok1, tt] > active > mark


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(52) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) Obligation:

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

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(s(X)) → PROPER(X)
PROPER(length(X)) → PROPER(X)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(isNat(X)) → PROPER(X)
PROPER(isNatList(X)) → PROPER(X)
PROPER(isNatIList(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.


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(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(isNat(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)
cons(x1, x2)  =  cons(x1, x2)
U11(x1, x2)  =  U11(x1, x2)
s(x1)  =  x1
length(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  isNat(x1)
isNatList(x1)  =  x1
isNatIList(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
0  =  0
tt  =  tt
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
zeros > [ok, top] > active1 > cons2 > [PROPER1, isNat1]
zeros > [ok, top] > active1 > U112 > [PROPER1, isNat1]
zeros > [ok, top] > active1 > and2 > [PROPER1, isNat1]
zeros > [ok, top] > active1 > [0, nil]
zeros > [ok, top] > active1 > tt
proper1 > [ok, top] > active1 > cons2 > [PROPER1, isNat1]
proper1 > [ok, top] > active1 > U112 > [PROPER1, isNat1]
proper1 > [ok, top] > active1 > and2 > [PROPER1, isNat1]
proper1 > [ok, top] > active1 > [0, nil]
proper1 > [ok, top] > active1 > tt


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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:

PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → PROPER(X)
PROPER(isNatList(X)) → PROPER(X)
PROPER(isNatIList(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.


PROPER(isNatList(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)  =  x1
length(x1)  =  x1
isNatList(x1)  =  isNatList(x1)
isNatIList(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark
cons(x1, x2)  =  cons(x2)
0  =  0
U11(x1, x2)  =  x1
tt  =  tt
and(x1, x2)  =  x1
isNat(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > [PROPER1, isNatList1, proper1] > zeros
active1 > [PROPER1, isNatList1, proper1] > [cons1, ok1, top] > mark
active1 > [PROPER1, isNatList1, proper1] > tt > mark
active1 > [PROPER1, isNatList1, proper1] > nil
0 > [cons1, ok1, top] > mark
0 > tt > mark


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(59) Obligation:

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

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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) 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
isNatIList(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
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  isNat
isNatList(x1)  =  isNatList
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > [and2, proper1] > s1
active1 > [and2, proper1] > zeros
active1 > [and2, proper1] > 0
active1 > [and2, proper1] > U112
active1 > [and2, proper1] > isNat > isNatList > tt
active1 > [and2, proper1] > nil


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(61) Obligation:

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

PROPER(length(X)) → PROPER(X)
PROPER(isNatIList(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(isNatIList(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
length(x1)  =  x1
isNatIList(x1)  =  isNatIList(x1)
active(x1)  =  x1
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  x2
0  =  0
U11(x1, x2)  =  x2
tt  =  tt
s(x1)  =  x1
and(x1, x2)  =  x2
isNat(x1)  =  isNat(x1)
isNatList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
nil > 0 > isNatIList1
nil > [tt, isNat1] > ok > top > isNatIList1
proper1 > zeros > 0 > isNatIList1
proper1 > zeros > [tt, isNat1] > ok > top > isNatIList1


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(63) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(64) 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)  =  x2
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
s(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  isNat
isNatList(x1)  =  isNatList
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[active1, isNatList, proper1, top] > zeros > [0, nil]
[active1, isNatList, proper1, top] > zeros > ok
[active1, isNatList, proper1, top] > tt > [length1, U112] > [0, nil]
[active1, isNatList, proper1, top] > tt > [length1, U112] > ok
[active1, isNatList, proper1, top] > and2 > ok
[active1, isNatList, proper1, top] > isNat > ok


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(65) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(66) PisEmptyProof (EQUIVALENT transformation)

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

(67) TRUE

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(69) 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
s(x1)  =  s(x1)
length(x1)  =  x1
and(x1, x2)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark
0  =  0
tt  =  tt
isNat(x1)  =  x1
isNatList(x1)  =  isNatList
isNatIList(x1)  =  isNatIList
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[active1, 0] > [mark, proper1] > s1
[active1, 0] > [mark, proper1] > zeros
[active1, 0] > [mark, proper1] > isNatIList > isNatList > tt
nil > [mark, proper1] > s1
nil > [mark, proper1] > zeros
nil > [mark, proper1] > isNatIList > isNatList > tt
top > [mark, proper1] > s1
top > [mark, proper1] > zeros
top > [mark, proper1] > isNatIList > isNatList > tt


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(70) 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(length(X)) → ACTIVE(X)
ACTIVE(and(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(71) 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)
ACTIVE(and(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)
length(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
active(x1)  =  x1
zeros  =  zeros
mark(x1)  =  mark
0  =  0
tt  =  tt
s(x1)  =  s
isNat(x1)  =  isNat
isNatList(x1)  =  x1
isNatIList(x1)  =  isNatIList
nil  =  nil
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[ACTIVE1, cons2, and2, zeros, tt, isNatIList] > [mark, s, isNat, nil] > 0


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(72) Obligation:

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

ACTIVE(U11(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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(73) 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)
length(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark
cons(x1, x2)  =  x1
0  =  0
tt  =  tt
s(x1)  =  s(x1)
and(x1, x2)  =  x1
isNat(x1)  =  isNat(x1)
isNatList(x1)  =  x1
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > [ACTIVE1, U111] > [zeros, mark, s1, ok] > isNat1
active1 > [ACTIVE1, U111] > [zeros, mark, s1, ok] > top
active1 > tt > [zeros, mark, s1, ok] > isNat1
active1 > tt > [zeros, mark, s1, ok] > top
0 > tt > [zeros, mark, s1, ok] > isNat1
0 > tt > [zeros, mark, s1, ok] > top
nil > [zeros, mark, s1, ok] > isNat1
nil > [zeros, mark, s1, ok] > top


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(74) Obligation:

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

ACTIVE(length(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(75) 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)
length(x1)  =  length(x1)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  x2
0  =  0
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
s(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  isNat
isNatList(x1)  =  isNatList
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[active1, isNatList, proper1, top] > zeros > [0, nil]
[active1, isNatList, proper1, top] > zeros > ok
[active1, isNatList, proper1, top] > tt > [length1, U112] > [0, nil]
[active1, isNatList, proper1, top] > tt > [length1, U112] > ok
[active1, isNatList, proper1, top] > and2 > ok
[active1, isNatList, proper1, top] > isNat > ok


The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(76) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.

(77) PisEmptyProof (EQUIVALENT transformation)

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

(78) TRUE

(79) 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(s(length(L)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(length(V1))) → mark(isNatList(V1))
active(isNat(s(V1))) → mark(isNat(V1))
active(isNatIList(V)) → mark(isNatList(V))
active(isNatIList(zeros)) → mark(tt)
active(isNatIList(cons(V1, V2))) → mark(and(isNat(V1), isNatIList(V2)))
active(isNatList(nil)) → mark(tt)
active(isNatList(cons(V1, V2))) → mark(and(isNat(V1), isNatList(V2)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(and(isNatList(L), isNat(N)), L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
active(and(X1, X2)) → and(active(X1), X2)
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
and(mark(X1), X2) → mark(and(X1, X2))
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(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(isNatList(X)) → isNatList(proper(X))
proper(isNatIList(X)) → isNatIList(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(X))
isNatList(ok(X)) → ok(isNatList(X))
isNatIList(ok(X)) → ok(isNatIList(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.