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(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__from(X)) → ACTIVATE(X)
2ND(cons(X, X1)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
2ND(cons(X, X1)) → 2ND(cons1(X, activate(X1)))

The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
2ND(cons(X, X1)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
2ND(cons(X, X1)) → 2ND(cons1(X, activate(X1)))

The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n__from(x1)) = 4 + x_1   
POL(n__s(x1)) = 4 + (4)x_1   
POL(ACTIVATE(x1)) = (4)x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.