Termination w.r.t. Q proof of /tmp/tmp23b1Jp/kabasci01.trs

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 PisEmptyProof (⇔)

    ↳24 TRUE

  ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 PisEmptyProof (⇔)

    ↳31 TRUE

  ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 PisEmptyProof (⇔)

    ↳38 TRUE

  ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 PisEmptyProof (⇔)

    ↳45 TRUE

  ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 PisEmptyProof (⇔)

    ↳52 TRUE

  ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 PisEmptyProof (⇔)

    ↳59 TRUE

  ↳60 QDP

(0) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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(eq(s(X), s(Y))) → EQ(X, Y)
ACTIVE(inf(X)) → CONS(X, inf(s(X)))
ACTIVE(inf(X)) → INF(s(X))
ACTIVE(inf(X)) → S(X)
ACTIVE(take(s(X), cons(Y, L))) → CONS(Y, take(X, L))
ACTIVE(take(s(X), cons(Y, L))) → TAKE(X, L)
ACTIVE(length(cons(X, L))) → S(length(L))
ACTIVE(length(cons(X, L))) → LENGTH(L)
ACTIVE(inf(X)) → INF(active(X))
ACTIVE(inf(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → TAKE(active(X1), X2)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → TAKE(X1, active(X2))
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(length(X)) → LENGTH(active(X))
ACTIVE(length(X)) → ACTIVE(X)
INF(mark(X)) → INF(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
PROPER(eq(X1, X2)) → EQ(proper(X1), proper(X2))
PROPER(eq(X1, X2)) → PROPER(X1)
PROPER(eq(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(inf(X)) → INF(proper(X))
PROPER(inf(X)) → PROPER(X)
PROPER(cons(any(X1), X2)) → CONS(any(any(proper(X1))), any(proper(X2)))
PROPER(cons(any(X1), X2)) → ANY(any(proper(X1)))
PROPER(cons(any(X1), X2)) → ANY(proper(X1))
PROPER(cons(any(X1), X2)) → PROPER(X1)
PROPER(cons(any(X1), X2)) → ANY(proper(X2))
PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(take(X1, X2)) → TAKE(proper(X1), proper(X2))
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(length(X)) → LENGTH(proper(X))
PROPER(length(X)) → PROPER(X)
EQ(ok(X1), ok(X2)) → EQ(X1, X2)
S(ok(X)) → S(X)
INF(ok(X)) → INF(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
LENGTH(ok(X)) → LENGTH(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)
ANY(X) → S(X)
ANY(proper(X)) → ANY(any(any(X)))
ANY(proper(X)) → ANY(any(X))
ANY(proper(X)) → ANY(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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 10 SCCs with 26 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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.

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
eq(x1, x2) = eq(x1)
0 = 0
mark(x1) = mark(x1)
true = true
s(x1) = x1
false = false
inf(x1) = x1
cons(x1, x2) = x2
take(x1, x2) = take(x1, x2)
nil = nil
length(x1) = length(x1)
proper(x1) = proper(x1)
any(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[active₁, eq₁, 0, false, take₂, proper₁] > true
[active₁, eq₁, 0, false, take₂, proper₁] > nil
[active₁, eq₁, 0, false, take₂, proper₁] > length₁ > [CONS₁, ok₁]
[active₁, eq₁, 0, false, take₂, proper₁] > length₁ > mark₁

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(7) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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:

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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.

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(x1)
eq(x1, x2) = x2
0 = 0
mark(x1) = mark
true = true
s(x1) = x1
false = false
inf(x1) = x1
cons(x1, x2) = x1
take(x1, x2) = x2
nil = nil
length(x1) = x1
proper(x1) = proper(x1)
any(x1) = any(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

S₁ > [active₁, mark, false, top]
[true, proper₁] > ok₁ > [active₁, mark, false, top]
[true, proper₁] > nil > 0 > [active₁, mark, false, top]
[true, proper₁] > any₁ > [active₁, mark, false, top]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(12) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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:

ANY(proper(X)) → ANY(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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.

ANY(proper(X)) → ANY(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ANY(x1) = ANY(x1)
proper(x1) = proper(x1)
active(x1) = x1
eq(x1, x2) = eq
0 = 0
mark(x1) = mark
true = true
s(x1) = s
false = false
inf(x1) = inf
cons(x1, x2) = cons(x1)
take(x1, x2) = take(x1, x2)
nil = nil
length(x1) = length(x1)
ok(x1) = ok
any(x1) = any
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[proper₁, eq, any] > ANY₁
[proper₁, eq, any] > 0 > nil > [true, s, ok, top] > [mark, take₂]
[proper₁, eq, any] > false
[proper₁, eq, any] > inf > [true, s, ok, top] > [mark, take₂]
[proper₁, eq, any] > cons₁ > [true, s, ok, top] > [mark, take₂]
[proper₁, eq, any] > length₁ > [true, s, ok, top] > [mark, take₂]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(17) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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:

EQ(ok(X1), ok(X2)) → EQ(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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.

EQ(ok(X1), ok(X2)) → EQ(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2) = EQ(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
eq(x1, x2) = eq(x1)
0 = 0
mark(x1) = mark(x1)
true = true
s(x1) = x1
false = false
inf(x1) = x1
cons(x1, x2) = x2
take(x1, x2) = take(x1, x2)
nil = nil
length(x1) = length(x1)
proper(x1) = proper(x1)
any(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[active₁, eq₁, 0, false, take₂, proper₁] > true
[active₁, eq₁, 0, false, take₂, proper₁] > nil
[active₁, eq₁, 0, false, take₂, proper₁] > length₁ > [EQ₁, ok₁]
[active₁, eq₁, 0, false, take₂, proper₁] > length₁ > mark₁

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(22) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE

(25) 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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LENGTH(mark(X)) → LENGTH(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH(x1) = x1
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
eq(x1, x2) = eq(x1)
0 = 0
true = true
s(x1) = x1
false = false
inf(x1) = x1
cons(x1, x2) = x2
take(x1, x2) = take(x1, x2)
nil = nil
length(x1) = x1
proper(x1) = proper(x1)
any(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[active₁, eq₁, 0, take₂, nil, proper₁] > mark₁ > [false, top]
[active₁, eq₁, 0, take₂, nil, proper₁] > true > [false, top]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(27) 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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LENGTH(ok(X)) → LENGTH(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH(x1) = LENGTH(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
eq(x1, x2) = x2
0 = 0
mark(x1) = mark
true = true
s(x1) = x1
false = false
inf(x1) = x1
cons(x1, x2) = x1
take(x1, x2) = x2
nil = nil
length(x1) = x1
proper(x1) = proper(x1)
any(x1) = any(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

LENGTH₁ > [active₁, mark, false, top]
[true, proper₁] > ok₁ > [active₁, mark, false, top]
[true, proper₁] > nil > 0 > [active₁, mark, false, top]
[true, proper₁] > any₁ > [active₁, mark, false, top]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(29) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) Obligation:

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

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

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2) = TAKE(x1, x2)
mark(x1) = mark(x1)
ok(x1) = x1
active(x1) = active(x1)
eq(x1, x2) = eq
0 = 0
true = true
s(x1) = s
false = false
inf(x1) = inf(x1)
cons(x1, x2) = x1
take(x1, x2) = take(x1, x2)
nil = nil
length(x1) = length(x1)
proper(x1) = x1
any(x1) = any
top(x1) = top

Recursive Path Order [RPO].
Precedence:

TAKE₂ > false
[0, nil] > [active₁, true] > eq > [mark₁, inf₁, length₁] > false
[0, nil] > [active₁, true] > s > [mark₁, inf₁, length₁] > false
[0, nil] > [active₁, true] > take₂ > [mark₁, inf₁, length₁] > false
any > s > [mark₁, inf₁, length₁] > false
top > false

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(34) Obligation:

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

TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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.

TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2) = TAKE(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
eq(x1, x2) = eq(x1)
0 = 0
mark(x1) = mark(x1)
true = true
s(x1) = x1
false = false
inf(x1) = x1
cons(x1, x2) = x2
take(x1, x2) = take(x1, x2)
nil = nil
length(x1) = length(x1)
proper(x1) = proper(x1)
any(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[active₁, eq₁, 0, false, take₂, proper₁] > true
[active₁, eq₁, 0, false, take₂, proper₁] > nil
[active₁, eq₁, 0, false, take₂, proper₁] > length₁ > [TAKE₁, ok₁]
[active₁, eq₁, 0, false, take₂, proper₁] > length₁ > mark₁

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(36) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(37) PisEmptyProof (EQUIVALENT transformation)

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

(38) TRUE

(39) Obligation:

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

INF(ok(X)) → INF(X)
INF(mark(X)) → INF(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INF(mark(X)) → INF(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INF(x1) = x1
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
eq(x1, x2) = eq(x1)
0 = 0
true = true
s(x1) = x1
false = false
inf(x1) = x1
cons(x1, x2) = x2
take(x1, x2) = take(x1, x2)
nil = nil
length(x1) = x1
proper(x1) = proper(x1)
any(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

[active₁, eq₁, 0, take₂, nil, proper₁] > mark₁ > [false, top]
[active₁, eq₁, 0, take₂, nil, proper₁] > true > [false, top]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(41) Obligation:

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

INF(ok(X)) → INF(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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.

INF(ok(X)) → INF(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INF(x1) = INF(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
eq(x1, x2) = x2
0 = 0
mark(x1) = mark
true = true
s(x1) = x1
false = false
inf(x1) = x1
cons(x1, x2) = x1
take(x1, x2) = x2
nil = nil
length(x1) = x1
proper(x1) = proper(x1)
any(x1) = any(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

INF₁ > [active₁, mark, false, top]
[true, proper₁] > ok₁ > [active₁, mark, false, top]
[true, proper₁] > nil > 0 > [active₁, mark, false, top]
[true, proper₁] > any₁ > [active₁, mark, false, top]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(43) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(44) PisEmptyProof (EQUIVALENT transformation)

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

(45) TRUE

(46) Obligation:

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

PROPER(eq(X1, X2)) → PROPER(X2)
PROPER(eq(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(inf(X)) → PROPER(X)
PROPER(cons(any(X1), X2)) → PROPER(X1)
PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(length(X)) → PROPER(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(eq(X1, X2)) → PROPER(X2)
PROPER(eq(X1, X2)) → PROPER(X1)
PROPER(cons(any(X1), X2)) → PROPER(X1)
PROPER(cons(any(X1), X2)) → PROPER(X2)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
eq(x1, x2) = eq(x1, x2)
s(x1) = x1
inf(x1) = x1
cons(x1, x2) = cons(x1, x2)
any(x1) = x1
take(x1, x2) = take(x1, x2)
length(x1) = x1
active(x1) = active(x1)
0 = 0
mark(x1) = x1
true = true
false = false
nil = nil
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Recursive Path Order [RPO].
Precedence:

top > active₁ > eq₂ > true > [take₂, false, nil, ok]
top > active₁ > cons₂ > [take₂, false, nil, ok]
top > active₁ > 0 > true > [take₂, false, nil, ok]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(48) Obligation:

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

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

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(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.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
s(x1) = s(x1)
inf(x1) = inf(x1)
length(x1) = length(x1)
active(x1) = active(x1)
eq(x1, x2) = eq
0 = 0
mark(x1) = x1
true = true
false = false
cons(x1, x2) = cons(x2)
take(x1, x2) = take
nil = nil
proper(x1) = proper(x1)
ok(x1) = x1
any(x1) = any(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

eq > proper₁ > [PROPER₁, length₁, active₁, true] > [s₁, any₁] > [inf₁, cons₁] > [false, take]
eq > proper₁ > [PROPER₁, length₁, active₁, true] > 0 > nil > [false, take]
top > [false, take]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(50) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) Obligation:

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

ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(inf(X)) → ACTIVE(X)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(length(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
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)
take(x1, x2) = take(x1, x2)
inf(x1) = x1
length(x1) = length(x1)
active(x1) = active(x1)
eq(x1, x2) = eq
0 = 0
mark(x1) = x1
true = true
s(x1) = s
false = false
cons(x1, x2) = x1
nil = nil
proper(x1) = proper(x1)
ok(x1) = x1
any(x1) = any
top(x1) = top

Recursive Path Order [RPO].
Precedence:

top > active₁ > [ACTIVE₁, take₂] > [length₁, s, nil]
top > active₁ > eq > true > [length₁, s, nil]
top > active₁ > eq > false > [length₁, s, nil]
top > active₁ > 0 > true > [length₁, s, nil]
top > [proper₁, any] > [ACTIVE₁, take₂] > [length₁, s, nil]
top > [proper₁, any] > eq > true > [length₁, s, nil]
top > [proper₁, any] > eq > false > [length₁, s, nil]
top > [proper₁, any] > 0 > true > [length₁, s, nil]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(55) Obligation:

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

ACTIVE(inf(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(inf(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)
inf(x1) = inf(x1)
active(x1) = active(x1)
eq(x1, x2) = eq
0 = 0
mark(x1) = x1
true = true
s(x1) = s
false = false
cons(x1, x2) = cons
take(x1, x2) = x1
nil = nil
length(x1) = length
proper(x1) = x1
ok(x1) = x1
any(x1) = any
top(x1) = top

Recursive Path Order [RPO].
Precedence:

ACTIVE₁ > [s, nil]
[active₁, top] > [inf₁, 0, cons, length, any] > true > [s, nil]
[active₁, top] > [eq, false] > true > [s, nil]

The following usable rules [FROCOS05] were oriented:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

(57) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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

(58) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE

(60) 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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
active(inf(X)) → inf(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(length(X)) → length(active(X))
inf(mark(X)) → mark(inf(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
length(mark(X)) → mark(length(X))
proper(eq(X1, X2)) → eq(proper(X1), proper(X2))
proper(0) → ok(0)
proper(true) → ok(true)
proper(s(X)) → s(proper(X))
proper(false) → ok(false)
proper(inf(X)) → inf(proper(X))
proper(cons(any(X1), X2)) → cons(any(any(proper(X1))), any(proper(X2)))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(length(X)) → length(proper(X))
eq(ok(X1), ok(X2)) → ok(eq(X1, X2))
s(ok(X)) → ok(s(X))
inf(ok(X)) → ok(inf(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
any(X) → s(X)
any(proper(X)) → any(any(any(X)))

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