(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(X1, X2)) → cons(proper(X1), 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))

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(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(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)

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(X1, X2)) → cons(proper(X1), 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))

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 9 SCCs with 20 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(X1, X2)) → cons(proper(X1), 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))

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

Lexicographic Path Order [LPO].
Precedence:
top > active1 > [0, take1, nil, proper1] > true > [ok1, eq1, inf1, cons1] > 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(X1, X2)) → cons(proper(X1), 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))

(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(X1, X2)) → cons(proper(X1), 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))

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(X1, X2)) → cons(proper(X1), 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))

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

Lexicographic Path Order [LPO].
Precedence:
top > [eq1, false, inf1, proper1] > [0, true] > [S1, ok1, mark, nil, length1]
top > [eq1, false, inf1, proper1] > cons1 > [S1, ok1, mark, nil, length1]


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(X1, X2)) → cons(proper(X1), 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))

(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(X1, X2)) → cons(proper(X1), 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))

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:

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(X1, X2)) → cons(proper(X1), 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))

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.


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

Lexicographic Path Order [LPO].
Precedence:
top > active1 > [0, take1, nil, proper1] > true > [ok1, eq1, inf1, cons1] > 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(X1, X2)) → cons(proper(X1), 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))

(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(X1, X2)) → cons(proper(X1), 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))

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:

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(X1, X2)) → cons(proper(X1), 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))

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.


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

Lexicographic Path Order [LPO].
Precedence:
active1 > 0 > true > false
active1 > 0 > nil > [ok1, s1, take1] > LENGTH1 > false
active1 > [cons1, length1, proper1, top] > [ok1, s1, take1] > LENGTH1 > false
active1 > [cons1, length1, proper1, top] > true > 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(X1, X2)) → cons(proper(X1), 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))

(22) Obligation:

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

LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(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(X1, X2)) → cons(proper(X1), 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))

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.


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

Lexicographic Path Order [LPO].
Precedence:
top > [active1, inf1] > 0 > [mark1, nil, length1] > LENGTH1 > false
top > [active1, inf1] > 0 > true > false
top > [active1, inf1] > cons2 > take2 > [mark1, nil, length1] > LENGTH1 > 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(X1, X2)) → cons(proper(X1), 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))

(24) 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(X1, X2)) → cons(proper(X1), 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))

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:

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(X1, X2)) → cons(proper(X1), 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))

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.


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

Lexicographic Path Order [LPO].
Precedence:
proper1 > [eq2, true] > false > [ok1, inf1]
proper1 > [take1, nil] > [ok1, inf1]
proper1 > [take1, nil] > 0


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(X1, X2)) → cons(proper(X1), 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))

(29) 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)

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(X1, X2)) → cons(proper(X1), 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))

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.


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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
eq(x1, x2)  =  x2
0  =  0
true  =  true
s(x1)  =  s(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)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[active1, take2, proper1] > nil > 0 > true > [mark1, s1, false]
top > [mark1, s1, 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(X1, X2)) → cons(proper(X1), 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))

(31) Obligation:

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

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

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

(32) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  TAKE(x2)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
eq(x1, x2)  =  x1
0  =  0
true  =  true
s(x1)  =  x1
false  =  false
inf(x1)  =  inf(x1)
cons(x1, x2)  =  x2
take(x1, x2)  =  take(x1, x2)
nil  =  nil
length(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
TAKE1 > false
active1 > inf1 > mark1 > top > false
active1 > take2 > mark1 > top > false
active1 > take2 > nil > [0, true] > 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(X1, X2)) → cons(proper(X1), 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))

(33) 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(X1, X2)) → cons(proper(X1), 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))

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) 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(X1, X2)) → cons(proper(X1), 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))

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.


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

Lexicographic Path Order [LPO].
Precedence:
active1 > 0 > true > false
active1 > 0 > nil > [ok1, s1, take1] > INF1 > false
active1 > [cons1, length1, proper1, top] > [ok1, s1, take1] > INF1 > false
active1 > [cons1, length1, proper1, top] > true > 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(X1, X2)) → cons(proper(X1), 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))

(38) Obligation:

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

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(X1, X2)) → cons(proper(X1), 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))

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

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

Lexicographic Path Order [LPO].
Precedence:
top > [active1, inf1] > 0 > [mark1, nil, length1] > INF1 > false
top > [active1, inf1] > 0 > true > false
top > [active1, inf1] > cons2 > take2 > [mark1, nil, length1] > INF1 > 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(X1, X2)) → cons(proper(X1), 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))

(40) 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(X1, X2)) → cons(proper(X1), 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))

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

(41) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) 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(X1, X2)) → PROPER(X1)
PROPER(cons(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(X1, X2)) → cons(proper(X1), 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))

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.


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

Lexicographic Path Order [LPO].
Precedence:
top > [take2, active1, proper1] > eq2 > ok1 > [0, mark, false, nil]
top > [take2, active1, proper1] > cons2 > ok1 > [0, mark, false, nil]
top > [take2, active1, proper1] > true > ok1 > [0, mark, false, 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(X1, X2)) → cons(proper(X1), 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))

(45) 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(X1, X2)) → cons(proper(X1), 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))

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

(46) 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(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)  =  x1
length(x1)  =  length(x1)
active(x1)  =  x1
eq(x1, x2)  =  eq
0  =  0
mark(x1)  =  mark
true  =  true
false  =  false
cons(x1, x2)  =  cons
take(x1, x2)  =  take(x2)
nil  =  nil
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
PROPER1 > [eq, mark]
s1 > [length1, cons, take1, nil] > [eq, mark]
0 > [length1, cons, take1, nil] > [eq, mark]
0 > true > [eq, mark]
false > [eq, mark]
top > [eq, mark]


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(X1, X2)) → cons(proper(X1), 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))

(47) Obligation:

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

PROPER(inf(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(X1, X2)) → cons(proper(X1), 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))

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

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
[active1, cons2, take2, nil] > [inf1, mark1] > top
[active1, cons2, take2, nil] > eq > true
[active1, cons2, take2, nil] > 0 > true
[active1, cons2, take2, nil] > 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(X1, X2)) → cons(proper(X1), 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))

(49) 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(X1, X2)) → cons(proper(X1), 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))

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

(50) PisEmptyProof (EQUIVALENT transformation)

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

(51) TRUE

(52) 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(X1, X2)) → cons(proper(X1), 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))

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

(53) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
take(x1, x2)  =  take(x1, x2)
inf(x1)  =  x1
length(x1)  =  x1
active(x1)  =  x1
eq(x1, x2)  =  eq
0  =  0
mark(x1)  =  x1
true  =  true
s(x1)  =  s
false  =  false
cons(x1, x2)  =  cons
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[0, nil] > [eq, true] > [take2, s, false, cons, proper1]
top > [take2, s, false, cons, proper1]


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(X1, X2)) → cons(proper(X1), 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))

(54) Obligation:

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

ACTIVE(inf(X)) → ACTIVE(X)
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(X1, X2)) → cons(proper(X1), 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))

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
length1 > 0 > [eq, true, s, false, cons]
take1 > nil > 0 > [eq, true, s, false, cons]
top1 > [eq, true, s, false, cons]


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(X1, X2)) → cons(proper(X1), 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))

(56) 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(X1, X2)) → cons(proper(X1), 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))

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

(57) 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)  =  mark(x1)
true  =  true
s(x1)  =  x1
false  =  false
cons(x1, x2)  =  cons(x1, x2)
take(x1, x2)  =  take(x1, x2)
nil  =  nil
length(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
[active1, cons2, take2, nil] > [inf1, mark1] > top
[active1, cons2, take2, nil] > eq > true
[active1, cons2, take2, nil] > 0 > true
[active1, cons2, take2, nil] > 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(X1, X2)) → cons(proper(X1), 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))

(58) 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(X1, X2)) → cons(proper(X1), 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))

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

(59) PisEmptyProof (EQUIVALENT transformation)

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

(60) TRUE

(61) 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(X1, X2)) → cons(proper(X1), 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))

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