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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[0, nil] > [ok1, active1, cons1, length1, isNatList1, proper1] > zeros > tt
[0, nil] > [ok1, active1, cons1, length1, isNatList1, proper1] > top

Status:
cons1: [1]
ISNATILIST1: [1]
ok1: [1]
0: []
active1: [1]
tt: []
isNatList1: [1]
zeros: []
proper1: [1]
length1: [1]
top: []
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))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[0, nil] > [ok1, active1, cons1, length1, isNatList1, proper1] > zeros > tt
[0, nil] > [ok1, active1, cons1, length1, isNatList1, proper1] > top

Status:
cons1: [1]
ok1: [1]
0: []
ISNATLIST1: [1]
active1: [1]
tt: []
isNatList1: [1]
zeros: []
proper1: [1]
length1: [1]
top: []
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))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[0, nil] > [ok1, active1, cons1, length1, isNatList1, proper1] > zeros > tt
[0, nil] > [ok1, active1, cons1, length1, isNatList1, proper1] > top

Status:
cons1: [1]
ok1: [1]
0: []
ISNAT1: [1]
active1: [1]
tt: []
isNatList1: [1]
zeros: []
proper1: [1]
length1: [1]
top: []
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))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[cons2, U111, s1, and1, isNat1, isNatList1, proper1] > [AND1, ok1, zeros, tt, isNatIList1, top] > 0 > mark
[cons2, U111, s1, and1, isNat1, isNatList1, proper1] > nil > mark

Status:
AND1: [1]
ok1: [1]
0: []
cons2: [1,2]
tt: []
isNatList1: [1]
zeros: []
isNatIList1: [1]
mark: []
and1: [1]
s1: [1]
U111: [1]
proper1: [1]
isNat1: [1]
top: []
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))

(22) Obligation:

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

AND(mark(X1), X2) → AND(X1, X2)

The TRS R consists of the following rules:

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[active1, cons2, length1, isNatList1, top] > zeros > 0
[active1, cons2, length1, isNatList1, top] > zeros > tt > mark1
[active1, cons2, length1, isNatList1, top] > U112 > mark1
[active1, cons2, length1, isNatList1, top] > and2 > mark1
[active1, cons2, length1, isNatList1, top] > isNat1 > mark1
[active1, cons2, length1, isNatList1, top] > isNatIList1
nil > 0
nil > tt > mark1

Status:
mark1: [1]
and2: [2,1]
U112: [2,1]
0: []
cons2: [1,2]
active1: [1]
tt: []
isNatList1: [1]
AND2: [1,2]
zeros: []
isNatIList1: [1]
isNat1: [1]
length1: [1]
top: []
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))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
active1 > zeros > tt
active1 > 0 > tt
[proper, top] > ok1
[proper, top] > zeros > tt
[proper, top] > 0 > tt
[proper, top] > nil

Status:
LENGTH1: [1]
active1: [1]
tt: []
zeros: []
proper: []
ok1: [1]
top: []
0: []
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))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > 0
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > U112 > mark1
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > U112 > ok
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > isNatIList > isNat > tt
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > isNatIList > isNat > ok
top > [active1, length1, and2, isNatList, proper1] > nil > mark1
top > [active1, length1, and2, isNatList, proper1] > nil > 0
top > [active1, length1, and2, isNatList, proper1] > nil > tt
top > [active1, length1, and2, isNatList, proper1] > nil > ok

Status:
LENGTH1: [1]
isNatList: []
mark1: [1]
U112: [2,1]
and2: [1,2]
isNatIList: []
0: []
isNat: []
active1: [1]
cons2: [1,2]
tt: []
zeros: []
ok: []
proper1: [1]
length1: [1]
top: []
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)  =  S(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
U11(x1, x2)  =  x2
tt  =  tt
s(x1)  =  x1
length(x1)  =  x1
and(x1, x2)  =  x2
isNat(x1)  =  x1
isNatList(x1)  =  x1
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper
top(x1)  =  top

Lexicographic path order with status [LPO].
Quasi-Precedence:
active1 > zeros > tt
active1 > 0 > tt
[proper, top] > ok1
[proper, top] > zeros > tt
[proper, top] > 0 > tt
[proper, top] > nil

Status:
active1: [1]
tt: []
zeros: []
proper: []
ok1: [1]
top: []
0: []
nil: []
S1: [1]


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > 0
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > U112 > mark1
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > U112 > ok
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > isNatIList > isNat > tt
top > [active1, length1, and2, isNatList, proper1] > [zeros, cons2] > isNatIList > isNat > ok
top > [active1, length1, and2, isNatList, proper1] > nil > mark1
top > [active1, length1, and2, isNatList, proper1] > nil > 0
top > [active1, length1, and2, isNatList, proper1] > nil > tt
top > [active1, length1, and2, isNatList, proper1] > nil > ok

Status:
isNatList: []
mark1: [1]
U112: [2,1]
and2: [1,2]
isNatIList: []
0: []
isNat: []
active1: [1]
cons2: [1,2]
tt: []
zeros: []
ok: []
proper1: [1]
length1: [1]
top: []
nil: []
S1: [1]


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[cons2, U111, s1, and1, isNat1, isNatList1, proper1] > [U11^11, ok1, zeros, tt, isNatIList1, top] > 0 > mark
[cons2, U111, s1, and1, isNat1, isNatList1, proper1] > nil > mark

Status:
U11^11: [1]
ok1: [1]
0: []
cons2: [1,2]
tt: []
isNatList1: [1]
zeros: []
isNatIList1: [1]
mark: []
and1: [1]
s1: [1]
U111: [1]
proper1: [1]
isNat1: [1]
top: []
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))

(43) Obligation:

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

U111(mark(X1), X2) → U111(X1, X2)

The TRS R consists of the following rules:

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[active1, cons2, length1, isNatList1, top] > zeros > 0
[active1, cons2, length1, isNatList1, top] > zeros > tt > mark1
[active1, cons2, length1, isNatList1, top] > U112 > mark1
[active1, cons2, length1, isNatList1, top] > and2 > mark1
[active1, cons2, length1, isNatList1, top] > isNat1 > mark1
[active1, cons2, length1, isNatList1, top] > isNatIList1
nil > 0
nil > tt > mark1

Status:
mark1: [1]
and2: [2,1]
U112: [2,1]
0: []
cons2: [1,2]
active1: [1]
tt: []
U11^12: [1,2]
isNatList1: [1]
zeros: []
isNatIList1: [1]
isNat1: [1]
length1: [1]
top: []
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))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[cons2, U111, s1, and1, isNat1, isNatList1, proper1] > [CONS1, ok1, zeros, tt, isNatIList1, top] > 0 > mark
[cons2, U111, s1, and1, isNat1, isNatList1, proper1] > nil > mark

Status:
CONS1: [1]
ok1: [1]
0: []
cons2: [1,2]
tt: []
isNatList1: [1]
zeros: []
isNatIList1: [1]
mark: []
and1: [1]
s1: [1]
U111: [1]
proper1: [1]
isNat1: [1]
top: []
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))

(50) Obligation:

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

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[active1, cons2, length1, isNatList1, top] > zeros > 0
[active1, cons2, length1, isNatList1, top] > zeros > tt > mark1
[active1, cons2, length1, isNatList1, top] > U112 > mark1
[active1, cons2, length1, isNatList1, top] > and2 > mark1
[active1, cons2, length1, isNatList1, top] > isNat1 > mark1
[active1, cons2, length1, isNatList1, top] > isNatIList1
nil > 0
nil > tt > mark1

Status:
mark1: [1]
and2: [2,1]
U112: [2,1]
0: []
cons2: [1,2]
active1: [1]
tt: []
isNatList1: [1]
zeros: []
CONS2: [1,2]
isNatIList1: [1]
isNat1: [1]
length1: [1]
top: []
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))

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
zeros > cons2 > and2
zeros > [length1, tt] > and2
zeros > [length1, tt] > 0
top > [U112, isNatList1, active1] > cons2 > and2
top > [U112, isNatList1, active1] > [length1, tt] > and2
top > [U112, isNatList1, active1] > [length1, tt] > 0

Status:
PROPER1: [1]
U112: [2,1]
and2: [2,1]
0: []
cons2: [2,1]
active1: [1]
tt: []
isNatList1: [1]
zeros: []
length1: [1]
top: []
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))

(57) Obligation:

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

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
isNat1 > [tt, nil] > 0 > ok > [active1, and1, isNatList1]
zeros > [tt, nil] > 0 > ok > [active1, and1, isNatList1]
top > [active1, and1, isNatList1]

Status:
active1: [1]
tt: []
isNatList1: [1]
zeros: []
and1: [1]
ok: []
isNat1: [1]
top: []
0: []
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))

(59) Obligation:

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

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
0 > [active1, mark] > [s1, isNatIList1, tt, and1, proper1, ok1] > zeros
0 > [active1, mark] > [s1, isNatIList1, tt, and1, proper1, ok1] > nil
top > [active1, mark] > [s1, isNatIList1, tt, and1, proper1, ok1] > zeros
top > [active1, mark] > [s1, isNatIList1, tt, and1, proper1, ok1] > nil

Status:
ok1: [1]
0: []
active1: [1]
tt: []
zeros: []
isNatIList1: [1]
mark: []
s1: [1]
and1: [1]
proper1: [1]
top: []
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:
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.

(62) PisEmptyProof (EQUIVALENT transformation)

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

(63) TRUE

(64) 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.

(65) 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)
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)  =  x1
s(x1)  =  x1
length(x1)  =  length(x1)
and(x1, x2)  =  and(x1, x2)
active(x1)  =  x1
zeros  =  zeros
mark(x1)  =  mark
0  =  0
tt  =  tt
isNat(x1)  =  isNat
isNatList(x1)  =  isNatList(x1)
isNatIList(x1)  =  isNatIList(x1)
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Quasi-Precedence:
[tt, isNat, isNatList1, isNatIList1, proper1, ok] > length1 > ACTIVE1
[tt, isNat, isNatList1, isNatIList1, proper1, ok] > length1 > [zeros, mark] > top
[tt, isNat, isNatList1, isNatIList1, proper1, ok] > and2 > ACTIVE1
[tt, isNat, isNatList1, isNatIList1, proper1, ok] > and2 > [zeros, mark] > top
[tt, isNat, isNatList1, isNatIList1, proper1, ok] > 0 > [zeros, mark] > top
[tt, isNat, isNatList1, isNatIList1, proper1, ok] > nil

Status:
and2: [2,1]
0: []
ACTIVE1: [1]
isNat: []
tt: []
isNatList1: [1]
zeros: []
isNatIList1: [1]
mark: []
ok: []
proper1: [1]
length1: [1]
top: []
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))

(66) 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)

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.

(67) 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)
cons(x1, x2)  =  x1
s(x1)  =  x1
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  mark
0  =  0
tt  =  tt
length(x1)  =  length(x1)
and(x1, x2)  =  x1
isNat(x1)  =  x1
isNatList(x1)  =  x1
isNatIList(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Lexicographic path order with status [LPO].
Quasi-Precedence:
active1 > [U111, 0, length1, proper1, ok1, top] > [zeros, tt]
active1 > [U111, 0, length1, proper1, ok1, top] > mark
active1 > [U111, 0, length1, proper1, ok1, top] > nil

Status:
ok1: [1]
0: []
ACTIVE1: [1]
active1: [1]
tt: []
zeros: []
mark: []
U111: [1]
proper1: [1]
length1: [1]
top: []
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))

(68) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(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.

(69) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
zeros > cons1
zeros > [tt, length, isNat, isNatList, isNatIList] > 0
nil > [tt, length, isNat, isNatList, isNatIList] > 0

Status:
isNat: []
cons1: [1]
tt: []
isNatList: []
zeros: []
isNatIList: []
length: []
0: []
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))

(70) Obligation:

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

ACTIVE(s(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.

(71) 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)
s(x1)  =  s(x1)
active(x1)  =  active(x1)
zeros  =  zeros
mark(x1)  =  x1
cons(x1, x2)  =  x1
0  =  0
U11(x1, x2)  =  U11
tt  =  tt
length(x1)  =  length
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  isNat
isNatList(x1)  =  isNatList
isNatIList(x1)  =  isNatIList
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Quasi-Precedence:
[active1, isNat, isNatList, proper1] > [s1, U11] > [length, nil, ok] > 0
[active1, isNat, isNatList, proper1] > zeros
[active1, isNat, isNatList, proper1] > isNatIList > tt > [length, nil, ok] > 0
[active1, isNat, isNatList, proper1] > isNatIList > and2 > [length, nil, ok] > 0

Status:
isNatList: []
U11: []
isNatIList: []
and2: [1,2]
length: []
0: []
ACTIVE1: [1]
isNat: []
active1: [1]
tt: []
zeros: []
s1: [1]
ok: []
proper1: [1]
top: []
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))

(72) 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.

(73) PisEmptyProof (EQUIVALENT transformation)

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

(74) TRUE

(75) 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.