(0) Obligation:

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

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

length(nil) → 0
length(cons(X, L)) → s(length(L))
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))

The TRS R 2 is

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false

The signature Sigma is {eq, true, false}

(2) Obligation:

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

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

EQ(s(X), s(Y)) → EQ(X, Y)
INF(X) → INF(s(X))
TAKE(s(X), cons(Y, L)) → TAKE(X, L)
LENGTH(cons(X, L)) → LENGTH(L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs.

(6) Complex Obligation (AND)

(7) Obligation:

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

LENGTH(cons(X, L)) → LENGTH(L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(8) Obligation:

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

TAKE(s(X), cons(Y, L)) → TAKE(X, L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(9) Obligation:

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

INF(X) → INF(s(X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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

(10) Obligation:

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

EQ(s(X), s(Y)) → EQ(X, Y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

eq(x0, x1)
inf(x0)
take(0, x0)
take(s(x0), cons(x1, x2))
length(nil)
length(cons(x0, x1))

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