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

(0) Obligation:

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

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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:

2ND(cons(X, n__cons(Y, Z))) → ACTIVATE(Y)
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
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)

The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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 1 SCC with 5 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(5) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
n__from(x1)  =  n__from(x1)
n__s(x1)  =  n__s(x1)
n__cons(x1, x2)  =  n__cons(x1)

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

• ACTIVATE(n__from(X)) → ACTIVATE(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• ACTIVATE(n__s(X)) → ACTIVATE(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(6) TRUE