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

(0) Obligation:

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

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

(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:

FROM(X) → FROM(s(X))
LENGTH(cons(X, Y)) → LENGTH1(Y)
LENGTH1(X) → LENGTH(X)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(6) Complex Obligation (AND)

(7) Obligation:

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

LENGTH1(X) → LENGTH(X)
LENGTH(cons(X, Y)) → LENGTH1(Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LENGTH1(X) → LENGTH(X)
LENGTH(cons(X, Y)) → LENGTH1(Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH1(x1)  =  LENGTH1(x1)
LENGTH(x1)  =  LENGTH(x1)
cons(x1, x2)  =  cons(x2)

Recursive path order with status [RPO].
Precedence:

cons₁ > LENGTH1₁ > LENGTH₁

Status:

LENGTH₁: multiset
LENGTH1₁: multiset
cons₁: multiset

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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