Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → 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:

FROM(X) → CONS(X, n__from(n__s(X)))
LENGTH(n__cons(X, Y)) → S(length1(activate(Y)))
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH(n__cons(X, Y)) → ACTIVATE(Y)
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → 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 2 SCCs with 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• ACTIVATE(n__s(X)) → ACTIVATE(X)
  The graph contains the following edges 1 > 1

• ACTIVATE(n__from(X)) → ACTIVATE(X)
  The graph contains the following edges 1 > 1

• ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
  The graph contains the following edges 1 > 1

(7) YES

(8) Obligation:

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(X) → LENGTH(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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

(9) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x1 / x0, x0 / n__s(x0)] on the rule
LENGTH(n__cons(x0, n__from(n__s(x0))))[ ]ⁿ[ ] → LENGTH(n__cons(x0, n__from(n__s(x0))))[ ]ⁿ[x1 / x0, x0 / n__s(x0)]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu
LENGTH(n__cons(x1, n__from(x0)))[ ]ⁿ[ ] → LENGTH(n__cons(x0, n__from(n__s(x0))))[ ]ⁿ[ ]
    by Rewrite t
        LENGTH(n__cons(x1, n__from(x0)))[ ]ⁿ[ ] → LENGTH(activate(from(activate(x0))))[ ]ⁿ[ ]
            by Narrowing at position: []
                intermediate steps: Instantiation
                LENGTH(n__cons(x1, n__from(y0)))[ ]ⁿ[ ] → LENGTH1(from(activate(y0)))[ ]ⁿ[ ]
                    by Narrowing at position: [0]
                        intermediate steps: Instantiation - Instantiation
                        LENGTH(n__cons(X, Y))[ ]ⁿ[ ] → LENGTH1(activate(Y))[ ]ⁿ[ ]
                            by OriginalRule from TRS P

                        intermediate steps: Instantiation
                        activate(n__from(X))[ ]ⁿ[ ] → from(activate(X))[ ]ⁿ[ ]
                            by OriginalRule from TRS R

                intermediate steps: Instantiation - Instantiation
                LENGTH1(X)[ ]ⁿ[ ] → LENGTH(activate(X))[ ]ⁿ[ ]
                    by OriginalRule from TRS P

(10) NO