Termination w.r.t. Q proof of /tmp/tmpcSRacY/LengthOfFiniteLists_nokinds

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 PisEmptyProof (⇔)

    ↳19 TRUE

  ↳20 QDP

    ↳21 QDPOrderProof (⇔)

    ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 PisEmptyProof (⇔)

    ↳33 TRUE

  ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 PisEmptyProof (⇔)

    ↳54 TRUE

  ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 QDPOrderProof (⇔)

    ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 PisEmptyProof (⇔)

    ↳63 TRUE

  ↳64 QDP

    ↳65 QDPOrderProof (⇔)

    ↳66 QDP

    ↳67 QDPOrderProof (⇔)

    ↳68 QDP

    ↳69 QDPOrderProof (⇔)

    ↳70 QDP

    ↳71 PisEmptyProof (⇔)

    ↳72 TRUE

  ↳73 QDP

(0) Obligation:

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

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

Lexicographic Path Order [LPO].
Precedence:

ISNATILIST₁ > [zeros, mark]
top > [nil, proper₁] > ok₁ > [active, 0, tt] > [zeros, mark]

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

ISNATLIST(ok(X)) → ISNATLIST(X)

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ISNATLIST(ok(X)) → ISNATLIST(X)

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

Lexicographic Path Order [LPO].
Precedence:

ISNATLIST₁ > [zeros, mark]
top > [nil, proper₁] > ok₁ > [active, 0, tt] > [zeros, mark]

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

ISNAT(ok(X)) → ISNAT(X)

The TRS R consists of the following rules:

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ISNAT(ok(X)) → ISNAT(X)

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

Lexicographic Path Order [LPO].
Precedence:

ISNAT₁ > [zeros, mark]
top > [nil, proper₁] > ok₁ > [active, 0, tt] > [zeros, mark]

The following usable rules [FROCOS05] were oriented:

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

(17) Obligation:

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

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

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(ok(X1), ok(X2)) → AND(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

zeros > [active₁, cons₁, 0, isNat₁, isNatIList₁, proper₁] > [ok₁, mark, tt, top]
nil > [active₁, cons₁, 0, isNat₁, isNatIList₁, proper₁] > [ok₁, mark, tt, top]

The following usable rules [FROCOS05] were oriented:

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

(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 [LPO].
Precedence:

zeros > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > U11₂ > [AND₂, mark₁, top]
zeros > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > and₂ > [AND₂, mark₁, top]
zeros > [active₁, tt, isNat₁, nil] > 0 > [AND₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > U11₂ > [AND₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > and₂ > [AND₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > 0 > [AND₂, mark₁, top]

The following usable rules [FROCOS05] were oriented:

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

(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(mark(X)) → LENGTH(X)

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

Lexicographic Path Order [LPO].
Precedence:

[zeros, proper₁] > [active₁, cons₂] > U11₂ > length₁ > [LENGTH₁, mark₁, s₁, isNatIList₁]
[zeros, proper₁] > [active₁, cons₂] > tt > length₁ > [LENGTH₁, mark₁, s₁, isNatIList₁]
[zeros, proper₁] > [active₁, cons₂] > and₂ > [LENGTH₁, mark₁, s₁, isNatIList₁]
[zeros, proper₁] > [0, nil] > tt > length₁ > [LENGTH₁, mark₁, s₁, isNatIList₁]
top > [active₁, cons₂] > U11₂ > length₁ > [LENGTH₁, mark₁, s₁, isNatIList₁]
top > [active₁, cons₂] > tt > length₁ > [LENGTH₁, mark₁, s₁, isNatIList₁]
top > [active₁, cons₂] > and₂ > [LENGTH₁, mark₁, s₁, isNatIList₁]

The following usable rules [FROCOS05] were oriented:

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

(29) Obligation:

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

LENGTH(ok(X)) → LENGTH(X)

The TRS R consists of the following rules:

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

Lexicographic Path Order [LPO].
Precedence:

LENGTH₁ > [zeros, mark]
top > [nil, proper₁] > ok₁ > [active, 0, tt] > [zeros, mark]

The following usable rules [FROCOS05] were oriented:

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

(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(mark(X)) → S(X)

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

Lexicographic Path Order [LPO].
Precedence:

[zeros, proper₁] > [active₁, cons₂] > U11₂ > length₁ > [S₁, mark₁, s₁, isNatIList₁]
[zeros, proper₁] > [active₁, cons₂] > tt > length₁ > [S₁, mark₁, s₁, isNatIList₁]
[zeros, proper₁] > [active₁, cons₂] > and₂ > [S₁, mark₁, s₁, isNatIList₁]
[zeros, proper₁] > [0, nil] > tt > length₁ > [S₁, mark₁, s₁, isNatIList₁]
top > [active₁, cons₂] > U11₂ > length₁ > [S₁, mark₁, s₁, isNatIList₁]
top > [active₁, cons₂] > tt > length₁ > [S₁, mark₁, s₁, isNatIList₁]
top > [active₁, cons₂] > and₂ > [S₁, mark₁, s₁, isNatIList₁]

The following usable rules [FROCOS05] were oriented:

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

(36) Obligation:

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

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

Lexicographic Path Order [LPO].
Precedence:

S₁ > [zeros, mark]
top > [nil, proper₁] > ok₁ > [active, 0, tt] > [zeros, mark]

The following usable rules [FROCOS05] were oriented:

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

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

Lexicographic Path Order [LPO].
Precedence:

zeros > [active₁, cons₁, 0, isNat₁, isNatIList₁, proper₁] > [ok₁, mark, tt, top]
nil > [active₁, cons₁, 0, isNat₁, isNatIList₁, proper₁] > [ok₁, mark, tt, top]

The following usable rules [FROCOS05] were oriented:

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

(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 [LPO].
Precedence:

zeros > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > U11₂ > [U11^1₂, mark₁, top]
zeros > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > and₂ > [U11^1₂, mark₁, top]
zeros > [active₁, tt, isNat₁, nil] > 0 > [U11^1₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > U11₂ > [U11^1₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > and₂ > [U11^1₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > 0 > [U11^1₂, mark₁, top]

The following usable rules [FROCOS05] were oriented:

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

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

Lexicographic Path Order [LPO].
Precedence:

zeros > [active₁, cons₁, 0, isNat₁, isNatIList₁, proper₁] > [ok₁, mark, tt, top]
nil > [active₁, cons₁, 0, isNat₁, isNatIList₁, proper₁] > [ok₁, mark, tt, top]

The following usable rules [FROCOS05] were oriented:

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

(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 [LPO].
Precedence:

zeros > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > U11₂ > [CONS₂, mark₁, top]
zeros > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > and₂ > [CONS₂, mark₁, top]
zeros > [active₁, tt, isNat₁, nil] > 0 > [CONS₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > U11₂ > [CONS₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > cons₂ > [length₁, isNatList₁] > and₂ > [CONS₂, mark₁, top]
isNatIList₁ > [active₁, tt, isNat₁, nil] > 0 > [CONS₂, mark₁, top]

The following usable rules [FROCOS05] were oriented:

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

(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(isNat(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) = 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) = isNat(x1)
isNatList(x1) = x1
isNatIList(x1) = isNatIList(x1)
active(x1) = x1
zeros = zeros
mark(x1) = mark
0 = 0
tt = tt
nil = nil
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

PROPER₁ > mark
[0, proper₁] > cons₂ > U11₂ > mark
[0, proper₁] > cons₂ > isNatIList₁ > and₂ > mark
[0, proper₁] > cons₂ > isNatIList₁ > [zeros, tt] > mark
[0, proper₁] > [length₁, isNat₁] > U11₂ > mark
[0, proper₁] > [length₁, isNat₁] > and₂ > mark
[0, proper₁] > [length₁, isNat₁] > [zeros, tt] > mark
[0, proper₁] > nil > [zeros, tt] > mark
top > mark

The following usable rules [FROCOS05] were oriented:

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

(57) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(isNatList(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(s(X)) → PROPER(X)

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

Lexicographic Path Order [LPO].
Precedence:

s₁ > PROPER₁ > [mark, cons₁, isNatIList₁]
s₁ > isNat > [mark, cons₁, isNatIList₁]
zeros > 0 > [mark, cons₁, isNatIList₁]
tt > [mark, cons₁, isNatIList₁]
length > 0 > [mark, cons₁, isNatIList₁]
length > and₂ > [mark, cons₁, isNatIList₁]
length > isNat > [mark, cons₁, isNatIList₁]
nil > 0 > [mark, cons₁, isNatIList₁]
top > [mark, cons₁, isNatIList₁]

The following usable rules [FROCOS05] were oriented:

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

(59) Obligation:

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

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

Lexicographic Path Order [LPO].
Precedence:

zeros > [0, proper₁, top] > isNatIList > ok > [active, mark] > [PROPER₁, isNatList₁]
zeros > [0, proper₁, top] > isNatIList > ok > [active, mark] > tt
zeros > [0, proper₁, 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(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → ACTIVE(X1)

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

Lexicographic Path Order [LPO].
Precedence:

active₁ > [zeros, tt] > 0 > [and₂, mark] > [cons₁, isNat] > [nil, ok]
active₁ > isNatIList > proper₁ > [and₂, mark] > [cons₁, isNat] > [nil, ok]
top > proper₁ > [and₂, mark] > [cons₁, isNat] > [nil, ok]

The following usable rules [FROCOS05] were oriented:

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

(66) Obligation:

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

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(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(s(X)) → ACTIVE(X)

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

Lexicographic Path Order [LPO].
Precedence:

tt > [mark, cons₁, isNatList₁] > [s₁, active₁] > ok > and₂
top > [isNat₁, isNatIList₁, proper₁] > nil > [zeros, 0] > [mark, cons₁, isNatList₁] > [s₁, active₁] > ok > and₂

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

The TRS R consists of the following rules:

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

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

(69) 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)
ACTIVE(length(X)) → ACTIVE(X)

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

Lexicographic Path Order [LPO].
Precedence:

U11₁ > length₁ > ACTIVE₁
U11₁ > length₁ > isNatList > tt > [mark, and₁]
zeros > tt > [mark, and₁]
cons₁ > isNatIList₁ > tt > [mark, and₁]
0 > tt > [mark, and₁]
isNat₁ > isNatList > tt > [mark, and₁]
nil > tt > [mark, and₁]

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:
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.

(71) PisEmptyProof (EQUIVALENT transformation)

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

(72) TRUE

(73) 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.