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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 PisEmptyProof (⇔)
↳8 TRUE

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__from(x1)  =  n__from(x1)
n__cons(x1, x2)  =  n__cons(x1)
n__s(x1)  =  n__s(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
[ACTIVATE1, nfrom1, ncons1]

Status:
nfrom1: multiset
ncons1: multiset
ns1: [1]
ACTIVATE1: [1]


The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

Q DP problem:
P is empty.
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.

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE