Termination w.r.t. Q of the following Term Rewriting System could be proven:

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(s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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(s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 4 less nodes.