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:

2nd1(cons2(X, n__cons2(Y, Z))) -> activate1(Y)
from1(X) -> cons2(X, n__from1(n__s1(X)))
cons2(X1, X2) -> n__cons2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

2nd1(cons2(X, n__cons2(Y, Z))) -> activate1(Y)
from1(X) -> cons2(X, n__from1(n__s1(X)))
cons2(X1, X2) -> n__cons2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(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:

ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
2ND1(cons2(X, n__cons2(Y, Z))) -> ACTIVATE1(Y)
FROM1(X) -> CONS2(X, n__from1(n__s1(X)))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(activate1(X1), X2)

The TRS R consists of the following rules:

2nd1(cons2(X, n__cons2(Y, Z))) -> activate1(Y)
from1(X) -> cons2(X, n__from1(n__s1(X)))
cons2(X1, X2) -> n__cons2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(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:

ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
2ND1(cons2(X, n__cons2(Y, Z))) -> ACTIVATE1(Y)
FROM1(X) -> CONS2(X, n__from1(n__s1(X)))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(activate1(X1), X2)

The TRS R consists of the following rules:

2nd1(cons2(X, n__cons2(Y, Z))) -> activate1(Y)
from1(X) -> cons2(X, n__from1(n__s1(X)))
cons2(X1, X2) -> n__cons2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(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 1 SCC with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

2nd1(cons2(X, n__cons2(Y, Z))) -> activate1(Y)
from1(X) -> cons2(X, n__from1(n__s1(X)))
cons2(X1, X2) -> n__cons2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE1(x1)) = x1   
POL(n__cons2(x1, x2)) = 1 + x1   
POL(n__from1(x1)) = 1 + x1   
POL(n__s1(x1)) = 1 + x1   

The following usable rules [14] 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:

2nd1(cons2(X, n__cons2(Y, Z))) -> activate1(Y)
from1(X) -> cons2(X, n__from1(n__s1(X)))
cons2(X1, X2) -> n__cons2(X1, X2)
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(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.